Prove that $arcsin z = frac{pi}{2} - arccos z$












1












$begingroup$


I have $arccos (z) = -iln (z + sqrt{z^2-1})$ and $arcsin (z)=-i ln(iz +sqrt{1-z^2}).$



Now I must prove, that $arcsin (z) = frac{pi}{2} - arccos (z)$.



I get:



$$arcsin (z)=-ilnleft(iz+sqrt{1-z^2}right)=-ilnleft(iz+sqrt{-1(z^2-1)}right)=-ilnleft(iz+sqrt{i^{2}(z^2-1)}right)=$$
$$=-lnleft(iz+isqrt{z^2-1}right)=-ilnleft(i(z+sqrt{z^2-1})right)=-iln -ilnleft(z+sqrt{z^2-1}right)=$$
$$=-ileft(ln|i|+ifrac{pi}{2}right)-ilnleft(z+sqrt{z^2-1}right)=frac{pi}{2}+arccos (z).$$



I don't know where I make mistake, because I have "+" not "-".






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
    $endgroup$
    – gammatester
    May 7 '14 at 14:19












  • $begingroup$
    so what I do now
    $endgroup$
    – user119543
    May 7 '14 at 14:26










  • $begingroup$
    Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
    $endgroup$
    – Daniel Fischer♦
    May 7 '14 at 14:27










  • $begingroup$
    but it is multiplicative funkction ... i would do made in this way ..
    $endgroup$
    – user119543
    May 7 '14 at 14:29










  • $begingroup$
    someone can help me?
    $endgroup$
    – user119543
    May 7 '14 at 14:46
















1












$begingroup$


I have $arccos (z) = -iln (z + sqrt{z^2-1})$ and $arcsin (z)=-i ln(iz +sqrt{1-z^2}).$



Now I must prove, that $arcsin (z) = frac{pi}{2} - arccos (z)$.



I get:



$$arcsin (z)=-ilnleft(iz+sqrt{1-z^2}right)=-ilnleft(iz+sqrt{-1(z^2-1)}right)=-ilnleft(iz+sqrt{i^{2}(z^2-1)}right)=$$
$$=-lnleft(iz+isqrt{z^2-1}right)=-ilnleft(i(z+sqrt{z^2-1})right)=-iln -ilnleft(z+sqrt{z^2-1}right)=$$
$$=-ileft(ln|i|+ifrac{pi}{2}right)-ilnleft(z+sqrt{z^2-1}right)=frac{pi}{2}+arccos (z).$$



I don't know where I make mistake, because I have "+" not "-".






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
    $endgroup$
    – gammatester
    May 7 '14 at 14:19












  • $begingroup$
    so what I do now
    $endgroup$
    – user119543
    May 7 '14 at 14:26










  • $begingroup$
    Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
    $endgroup$
    – Daniel Fischer♦
    May 7 '14 at 14:27










  • $begingroup$
    but it is multiplicative funkction ... i would do made in this way ..
    $endgroup$
    – user119543
    May 7 '14 at 14:29










  • $begingroup$
    someone can help me?
    $endgroup$
    – user119543
    May 7 '14 at 14:46














1












1








1


1



$begingroup$


I have $arccos (z) = -iln (z + sqrt{z^2-1})$ and $arcsin (z)=-i ln(iz +sqrt{1-z^2}).$



Now I must prove, that $arcsin (z) = frac{pi}{2} - arccos (z)$.



I get:



$$arcsin (z)=-ilnleft(iz+sqrt{1-z^2}right)=-ilnleft(iz+sqrt{-1(z^2-1)}right)=-ilnleft(iz+sqrt{i^{2}(z^2-1)}right)=$$
$$=-lnleft(iz+isqrt{z^2-1}right)=-ilnleft(i(z+sqrt{z^2-1})right)=-iln -ilnleft(z+sqrt{z^2-1}right)=$$
$$=-ileft(ln|i|+ifrac{pi}{2}right)-ilnleft(z+sqrt{z^2-1}right)=frac{pi}{2}+arccos (z).$$



I don't know where I make mistake, because I have "+" not "-".






complex-analysis







share|cite|improve this question











$endgroup$




I have $arccos (z) = -iln (z + sqrt{z^2-1})$ and $arcsin (z)=-i ln(iz +sqrt{1-z^2}).$



Now I must prove, that $arcsin (z) = frac{pi}{2} - arccos (z)$.



I get:



$$arcsin (z)=-ilnleft(iz+sqrt{1-z^2}right)=-ilnleft(iz+sqrt{-1(z^2-1)}right)=-ilnleft(iz+sqrt{i^{2}(z^2-1)}right)=$$
$$=-lnleft(iz+isqrt{z^2-1}right)=-ilnleft(i(z+sqrt{z^2-1})right)=-iln -ilnleft(z+sqrt{z^2-1}right)=$$
$$=-ileft(ln|i|+ifrac{pi}{2}right)-ilnleft(z+sqrt{z^2-1}right)=frac{pi}{2}+arccos (z).$$



I don't know where I make mistake, because I have "+" not "-".






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 7 '14 at 14:23









Samrat Mukhopadhyay

13.7k2047




13.7k2047










asked May 7 '14 at 14:11









user119543user119543

115




115












  • $begingroup$
    One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
    $endgroup$
    – gammatester
    May 7 '14 at 14:19












  • $begingroup$
    so what I do now
    $endgroup$
    – user119543
    May 7 '14 at 14:26










  • $begingroup$
    Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
    $endgroup$
    – Daniel Fischer♦
    May 7 '14 at 14:27










  • $begingroup$
    but it is multiplicative funkction ... i would do made in this way ..
    $endgroup$
    – user119543
    May 7 '14 at 14:29










  • $begingroup$
    someone can help me?
    $endgroup$
    – user119543
    May 7 '14 at 14:46


















  • $begingroup$
    One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
    $endgroup$
    – gammatester
    May 7 '14 at 14:19












  • $begingroup$
    so what I do now
    $endgroup$
    – user119543
    May 7 '14 at 14:26










  • $begingroup$
    Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
    $endgroup$
    – Daniel Fischer♦
    May 7 '14 at 14:27










  • $begingroup$
    but it is multiplicative funkction ... i would do made in this way ..
    $endgroup$
    – user119543
    May 7 '14 at 14:29










  • $begingroup$
    someone can help me?
    $endgroup$
    – user119543
    May 7 '14 at 14:46
















$begingroup$
One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
$endgroup$
– gammatester
May 7 '14 at 14:19






$begingroup$
One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
$endgroup$
– gammatester
May 7 '14 at 14:19














$begingroup$
so what I do now
$endgroup$
– user119543
May 7 '14 at 14:26




$begingroup$
so what I do now
$endgroup$
– user119543
May 7 '14 at 14:26












$begingroup$
Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
$endgroup$
– Daniel Fischer♦
May 7 '14 at 14:27




$begingroup$
Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
$endgroup$
– Daniel Fischer♦
May 7 '14 at 14:27












$begingroup$
but it is multiplicative funkction ... i would do made in this way ..
$endgroup$
– user119543
May 7 '14 at 14:29




$begingroup$
but it is multiplicative funkction ... i would do made in this way ..
$endgroup$
– user119543
May 7 '14 at 14:29












$begingroup$
someone can help me?
$endgroup$
– user119543
May 7 '14 at 14:46




$begingroup$
someone can help me?
$endgroup$
– user119543
May 7 '14 at 14:46










3 Answers
3






active

oldest

votes


















0












$begingroup$

Suppose the statement is true, namely,
begin{align}
sin^{-1}(z) = frac{pi}{2} - cos^{-1}(z).
end{align}
By taking the sine of both sides leads to
begin{align}
sin(sin^{-1}(z)) &= sin(frac{pi}{2} - cos^{-1}(z)) \
z &= sin(pi/2) cos(cos^{-1}(z)) - cos(pi/2) sin(cos^{-1}(z)) \
z &= z.
end{align}
This shows the identity is true.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    lets start by defining $w to be $sin(w) = z
    now, we'll use the complex identity:



    sin(w) = z = (exp(iw)-exp(-iw))/2i


    define t = exp(iw) therefor, by applying this to the previous equation:



    z = (t - 1/t)/2i


    multiply by 2i and pass all args to the right side to get:



    t^2 - (2iz) - 1 = 0


    a standard quadratic equation:



    t = ((2iz)+sqrt((2i*z)^2 + 4))/2
    
t = i*z + sqrt(1 - z^2)


    but we defined t = exp(i*w) therfor:



    exp(iw) = i*z + sqrt(1 - z^2)


    take the log to both sides and get that:



    iw = log(i*z + sqrt(1 - z^2))
    divide by i:



    w = (1/i) * log(i*z + sqrt(1 - z^2))


    but as we first defined: sin(w) = z thenw = arcsin(z)
    in total:



    arcsin(z) = (1/i) * log(i*z + sqrt(1 - z^2))


    that proves your claim is right.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Please do not bump old posts unless absolutely necessary.
      $endgroup$
      – Cyclohexanol.
      Dec 3 '14 at 23:14



















    0












    $begingroup$

    My usual naive approach.



    From
    $arccos (z) = -iln (z + sqrt{z^2-1})
    $
    and
    $arcsin (z)=-i ln(iz +sqrt{1-z^2})
    $
    we get



    $begin{array}\
    arccos (z)+arcsin (z)
    &=-i(ln (z + sqrt{z^2-1})+ln(iz +sqrt{1-z^2}))\
    &=-i(ln ((z + sqrt{z^2-1})(iz +sqrt{1-z^2}))\
    &=-i(ln ((z + isqrt{1-z^2})(iz +sqrt{1-z^2}))\
    &=-i(ln (iz^2+zsqrt{1-z^2}(-1+1)+i(1-z^2))\
    &=-i(ln (i))\
    &=-i(dfrac{ipi}{2})\
    &=dfrac{pi}{2}\
    end{array}
    $



    Throw in a $2kpi i$
    if you want.
    That will change the answer to
    $dfrac{pi}{2}+2kpi$.






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f785028%2fprove-that-arcsin-z-frac-pi2-arccos-z%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Suppose the statement is true, namely,
      begin{align}
      sin^{-1}(z) = frac{pi}{2} - cos^{-1}(z).
      end{align}
      By taking the sine of both sides leads to
      begin{align}
      sin(sin^{-1}(z)) &= sin(frac{pi}{2} - cos^{-1}(z)) \
      z &= sin(pi/2) cos(cos^{-1}(z)) - cos(pi/2) sin(cos^{-1}(z)) \
      z &= z.
      end{align}
      This shows the identity is true.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Suppose the statement is true, namely,
        begin{align}
        sin^{-1}(z) = frac{pi}{2} - cos^{-1}(z).
        end{align}
        By taking the sine of both sides leads to
        begin{align}
        sin(sin^{-1}(z)) &= sin(frac{pi}{2} - cos^{-1}(z)) \
        z &= sin(pi/2) cos(cos^{-1}(z)) - cos(pi/2) sin(cos^{-1}(z)) \
        z &= z.
        end{align}
        This shows the identity is true.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Suppose the statement is true, namely,
          begin{align}
          sin^{-1}(z) = frac{pi}{2} - cos^{-1}(z).
          end{align}
          By taking the sine of both sides leads to
          begin{align}
          sin(sin^{-1}(z)) &= sin(frac{pi}{2} - cos^{-1}(z)) \
          z &= sin(pi/2) cos(cos^{-1}(z)) - cos(pi/2) sin(cos^{-1}(z)) \
          z &= z.
          end{align}
          This shows the identity is true.






          share|cite|improve this answer









          $endgroup$



          Suppose the statement is true, namely,
          begin{align}
          sin^{-1}(z) = frac{pi}{2} - cos^{-1}(z).
          end{align}
          By taking the sine of both sides leads to
          begin{align}
          sin(sin^{-1}(z)) &= sin(frac{pi}{2} - cos^{-1}(z)) \
          z &= sin(pi/2) cos(cos^{-1}(z)) - cos(pi/2) sin(cos^{-1}(z)) \
          z &= z.
          end{align}
          This shows the identity is true.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 7 '14 at 17:39









          LeucippusLeucippus

          19.6k102871




          19.6k102871























              0












              $begingroup$

              lets start by defining $w to be $sin(w) = z
              now, we'll use the complex identity:



              sin(w) = z = (exp(iw)-exp(-iw))/2i


              define t = exp(iw) therefor, by applying this to the previous equation:



              z = (t - 1/t)/2i


              multiply by 2i and pass all args to the right side to get:



              t^2 - (2iz) - 1 = 0


              a standard quadratic equation:



              t = ((2iz)+sqrt((2i*z)^2 + 4))/2
              
t = i*z + sqrt(1 - z^2)


              but we defined t = exp(i*w) therfor:



              exp(iw) = i*z + sqrt(1 - z^2)


              take the log to both sides and get that:



              iw = log(i*z + sqrt(1 - z^2))
              divide by i:



              w = (1/i) * log(i*z + sqrt(1 - z^2))


              but as we first defined: sin(w) = z thenw = arcsin(z)
              in total:



              arcsin(z) = (1/i) * log(i*z + sqrt(1 - z^2))


              that proves your claim is right.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Please do not bump old posts unless absolutely necessary.
                $endgroup$
                – Cyclohexanol.
                Dec 3 '14 at 23:14
















              0












              $begingroup$

              lets start by defining $w to be $sin(w) = z
              now, we'll use the complex identity:



              sin(w) = z = (exp(iw)-exp(-iw))/2i


              define t = exp(iw) therefor, by applying this to the previous equation:



              z = (t - 1/t)/2i


              multiply by 2i and pass all args to the right side to get:



              t^2 - (2iz) - 1 = 0


              a standard quadratic equation:



              t = ((2iz)+sqrt((2i*z)^2 + 4))/2
              
t = i*z + sqrt(1 - z^2)


              but we defined t = exp(i*w) therfor:



              exp(iw) = i*z + sqrt(1 - z^2)


              take the log to both sides and get that:



              iw = log(i*z + sqrt(1 - z^2))
              divide by i:



              w = (1/i) * log(i*z + sqrt(1 - z^2))


              but as we first defined: sin(w) = z thenw = arcsin(z)
              in total:



              arcsin(z) = (1/i) * log(i*z + sqrt(1 - z^2))


              that proves your claim is right.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Please do not bump old posts unless absolutely necessary.
                $endgroup$
                – Cyclohexanol.
                Dec 3 '14 at 23:14














              0












              0








              0





              $begingroup$

              lets start by defining $w to be $sin(w) = z
              now, we'll use the complex identity:



              sin(w) = z = (exp(iw)-exp(-iw))/2i


              define t = exp(iw) therefor, by applying this to the previous equation:



              z = (t - 1/t)/2i


              multiply by 2i and pass all args to the right side to get:



              t^2 - (2iz) - 1 = 0


              a standard quadratic equation:



              t = ((2iz)+sqrt((2i*z)^2 + 4))/2
              
t = i*z + sqrt(1 - z^2)


              but we defined t = exp(i*w) therfor:



              exp(iw) = i*z + sqrt(1 - z^2)


              take the log to both sides and get that:



              iw = log(i*z + sqrt(1 - z^2))
              divide by i:



              w = (1/i) * log(i*z + sqrt(1 - z^2))


              but as we first defined: sin(w) = z thenw = arcsin(z)
              in total:



              arcsin(z) = (1/i) * log(i*z + sqrt(1 - z^2))


              that proves your claim is right.






              share|cite|improve this answer









              $endgroup$



              lets start by defining $w to be $sin(w) = z
              now, we'll use the complex identity:



              sin(w) = z = (exp(iw)-exp(-iw))/2i


              define t = exp(iw) therefor, by applying this to the previous equation:



              z = (t - 1/t)/2i


              multiply by 2i and pass all args to the right side to get:



              t^2 - (2iz) - 1 = 0


              a standard quadratic equation:



              t = ((2iz)+sqrt((2i*z)^2 + 4))/2
              
t = i*z + sqrt(1 - z^2)


              but we defined t = exp(i*w) therfor:



              exp(iw) = i*z + sqrt(1 - z^2)


              take the log to both sides and get that:



              iw = log(i*z + sqrt(1 - z^2))
              divide by i:



              w = (1/i) * log(i*z + sqrt(1 - z^2))


              but as we first defined: sin(w) = z thenw = arcsin(z)
              in total:



              arcsin(z) = (1/i) * log(i*z + sqrt(1 - z^2))


              that proves your claim is right.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 3 '14 at 23:09









              ThunderWiringThunderWiring

              1215




              1215












              • $begingroup$
                Please do not bump old posts unless absolutely necessary.
                $endgroup$
                – Cyclohexanol.
                Dec 3 '14 at 23:14


















              • $begingroup$
                Please do not bump old posts unless absolutely necessary.
                $endgroup$
                – Cyclohexanol.
                Dec 3 '14 at 23:14
















              $begingroup$
              Please do not bump old posts unless absolutely necessary.
              $endgroup$
              – Cyclohexanol.
              Dec 3 '14 at 23:14




              $begingroup$
              Please do not bump old posts unless absolutely necessary.
              $endgroup$
              – Cyclohexanol.
              Dec 3 '14 at 23:14











              0












              $begingroup$

              My usual naive approach.



              From
              $arccos (z) = -iln (z + sqrt{z^2-1})
              $
              and
              $arcsin (z)=-i ln(iz +sqrt{1-z^2})
              $
              we get



              $begin{array}\
              arccos (z)+arcsin (z)
              &=-i(ln (z + sqrt{z^2-1})+ln(iz +sqrt{1-z^2}))\
              &=-i(ln ((z + sqrt{z^2-1})(iz +sqrt{1-z^2}))\
              &=-i(ln ((z + isqrt{1-z^2})(iz +sqrt{1-z^2}))\
              &=-i(ln (iz^2+zsqrt{1-z^2}(-1+1)+i(1-z^2))\
              &=-i(ln (i))\
              &=-i(dfrac{ipi}{2})\
              &=dfrac{pi}{2}\
              end{array}
              $



              Throw in a $2kpi i$
              if you want.
              That will change the answer to
              $dfrac{pi}{2}+2kpi$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                My usual naive approach.



                From
                $arccos (z) = -iln (z + sqrt{z^2-1})
                $
                and
                $arcsin (z)=-i ln(iz +sqrt{1-z^2})
                $
                we get



                $begin{array}\
                arccos (z)+arcsin (z)
                &=-i(ln (z + sqrt{z^2-1})+ln(iz +sqrt{1-z^2}))\
                &=-i(ln ((z + sqrt{z^2-1})(iz +sqrt{1-z^2}))\
                &=-i(ln ((z + isqrt{1-z^2})(iz +sqrt{1-z^2}))\
                &=-i(ln (iz^2+zsqrt{1-z^2}(-1+1)+i(1-z^2))\
                &=-i(ln (i))\
                &=-i(dfrac{ipi}{2})\
                &=dfrac{pi}{2}\
                end{array}
                $



                Throw in a $2kpi i$
                if you want.
                That will change the answer to
                $dfrac{pi}{2}+2kpi$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  My usual naive approach.



                  From
                  $arccos (z) = -iln (z + sqrt{z^2-1})
                  $
                  and
                  $arcsin (z)=-i ln(iz +sqrt{1-z^2})
                  $
                  we get



                  $begin{array}\
                  arccos (z)+arcsin (z)
                  &=-i(ln (z + sqrt{z^2-1})+ln(iz +sqrt{1-z^2}))\
                  &=-i(ln ((z + sqrt{z^2-1})(iz +sqrt{1-z^2}))\
                  &=-i(ln ((z + isqrt{1-z^2})(iz +sqrt{1-z^2}))\
                  &=-i(ln (iz^2+zsqrt{1-z^2}(-1+1)+i(1-z^2))\
                  &=-i(ln (i))\
                  &=-i(dfrac{ipi}{2})\
                  &=dfrac{pi}{2}\
                  end{array}
                  $



                  Throw in a $2kpi i$
                  if you want.
                  That will change the answer to
                  $dfrac{pi}{2}+2kpi$.






                  share|cite|improve this answer









                  $endgroup$



                  My usual naive approach.



                  From
                  $arccos (z) = -iln (z + sqrt{z^2-1})
                  $
                  and
                  $arcsin (z)=-i ln(iz +sqrt{1-z^2})
                  $
                  we get



                  $begin{array}\
                  arccos (z)+arcsin (z)
                  &=-i(ln (z + sqrt{z^2-1})+ln(iz +sqrt{1-z^2}))\
                  &=-i(ln ((z + sqrt{z^2-1})(iz +sqrt{1-z^2}))\
                  &=-i(ln ((z + isqrt{1-z^2})(iz +sqrt{1-z^2}))\
                  &=-i(ln (iz^2+zsqrt{1-z^2}(-1+1)+i(1-z^2))\
                  &=-i(ln (i))\
                  &=-i(dfrac{ipi}{2})\
                  &=dfrac{pi}{2}\
                  end{array}
                  $



                  Throw in a $2kpi i$
                  if you want.
                  That will change the answer to
                  $dfrac{pi}{2}+2kpi$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 28 '18 at 21:49









                  marty cohenmarty cohen

                  73.2k549128




                  73.2k549128






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f785028%2fprove-that-arcsin-z-frac-pi2-arccos-z%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Human spaceflight

                      Image
                      Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
                      Read more

                      Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                      Image
                      0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
                      Read more

                      File:DeusFollowingSea.jpg

                      Image
                      File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
                      Read more