Find trigonometric form of complex number $frac{1}{2}-frac{1}{sqrt{2}}i$
$begingroup$
Find trigonometric form of complex number $frac{1}{2}-frac{1}{sqrt{2}}i$.
I assume that this is related to trigonometric form of complex number
$$1 pm cos(alpha) pm isin(alpha)$$ or similiar.
Is it possible to find such form without taking arctangent?
complex-numbers
$endgroup$
add a comment |
$begingroup$
Find trigonometric form of complex number $frac{1}{2}-frac{1}{sqrt{2}}i$.
I assume that this is related to trigonometric form of complex number
$$1 pm cos(alpha) pm isin(alpha)$$ or similiar.
Is it possible to find such form without taking arctangent?
complex-numbers
$endgroup$
1
$begingroup$
Why the $1$? Why not $r(cosalpha+isinalpha)$?
$endgroup$
– Lord Shark the Unknown
Jan 8 at 19:16
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hint: find modulus first then divide real part by modulus to get cosine, divide imaginary part by modulus to get sine
$endgroup$
– Vasya
Jan 8 at 19:16
1
$begingroup$
What's wrong with taking the arctangent? You can divide the result by $pi$ to so if the result is a nice fraction (it isn't). You can draw a picture and do some symilar triangle relationship chasing but I don't think you get any "nice" angles.
$endgroup$
– fleablood
Jan 8 at 19:50
$begingroup$
Our professor told us to do so, but I suppose there must be mistake if such "horrible" angles occur.
$endgroup$
– janusz
Jan 8 at 19:52
add a comment |
$begingroup$
Find trigonometric form of complex number $frac{1}{2}-frac{1}{sqrt{2}}i$.
I assume that this is related to trigonometric form of complex number
$$1 pm cos(alpha) pm isin(alpha)$$ or similiar.
Is it possible to find such form without taking arctangent?
complex-numbers
$endgroup$
Find trigonometric form of complex number $frac{1}{2}-frac{1}{sqrt{2}}i$.
I assume that this is related to trigonometric form of complex number
$$1 pm cos(alpha) pm isin(alpha)$$ or similiar.
Is it possible to find such form without taking arctangent?
complex-numbers
complex-numbers
edited Jan 8 at 19:22
janusz
asked Jan 8 at 19:14
januszjanusz
462210
462210
1
$begingroup$
Why the $1$? Why not $r(cosalpha+isinalpha)$?
$endgroup$
– Lord Shark the Unknown
Jan 8 at 19:16
$begingroup$
hint: find modulus first then divide real part by modulus to get cosine, divide imaginary part by modulus to get sine
$endgroup$
– Vasya
Jan 8 at 19:16
1
$begingroup$
What's wrong with taking the arctangent? You can divide the result by $pi$ to so if the result is a nice fraction (it isn't). You can draw a picture and do some symilar triangle relationship chasing but I don't think you get any "nice" angles.
$endgroup$
– fleablood
Jan 8 at 19:50
$begingroup$
Our professor told us to do so, but I suppose there must be mistake if such "horrible" angles occur.
$endgroup$
– janusz
Jan 8 at 19:52
add a comment |
1
$begingroup$
Why the $1$? Why not $r(cosalpha+isinalpha)$?
$endgroup$
– Lord Shark the Unknown
Jan 8 at 19:16
$begingroup$
hint: find modulus first then divide real part by modulus to get cosine, divide imaginary part by modulus to get sine
$endgroup$
– Vasya
Jan 8 at 19:16
1
$begingroup$
What's wrong with taking the arctangent? You can divide the result by $pi$ to so if the result is a nice fraction (it isn't). You can draw a picture and do some symilar triangle relationship chasing but I don't think you get any "nice" angles.
$endgroup$
– fleablood
Jan 8 at 19:50
$begingroup$
Our professor told us to do so, but I suppose there must be mistake if such "horrible" angles occur.
$endgroup$
– janusz
Jan 8 at 19:52
1
1
$begingroup$
Why the $1$? Why not $r(cosalpha+isinalpha)$?
$endgroup$
– Lord Shark the Unknown
Jan 8 at 19:16
$begingroup$
Why the $1$? Why not $r(cosalpha+isinalpha)$?
$endgroup$
– Lord Shark the Unknown
Jan 8 at 19:16
$begingroup$
hint: find modulus first then divide real part by modulus to get cosine, divide imaginary part by modulus to get sine
$endgroup$
– Vasya
Jan 8 at 19:16
$begingroup$
hint: find modulus first then divide real part by modulus to get cosine, divide imaginary part by modulus to get sine
$endgroup$
– Vasya
Jan 8 at 19:16
1
1
$begingroup$
What's wrong with taking the arctangent? You can divide the result by $pi$ to so if the result is a nice fraction (it isn't). You can draw a picture and do some symilar triangle relationship chasing but I don't think you get any "nice" angles.
$endgroup$
– fleablood
Jan 8 at 19:50
$begingroup$
What's wrong with taking the arctangent? You can divide the result by $pi$ to so if the result is a nice fraction (it isn't). You can draw a picture and do some symilar triangle relationship chasing but I don't think you get any "nice" angles.
$endgroup$
– fleablood
Jan 8 at 19:50
$begingroup$
Our professor told us to do so, but I suppose there must be mistake if such "horrible" angles occur.
$endgroup$
– janusz
Jan 8 at 19:52
$begingroup$
Our professor told us to do so, but I suppose there must be mistake if such "horrible" angles occur.
$endgroup$
– janusz
Jan 8 at 19:52
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Just do it.
$|frac 12 - frac 1{sqrt 2} i| = sqrt {frac 12^2 + frac 12} = frac {sqrt 3} 2$
So $frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2 (frac 1{sqrt 3} - frac {sqrt 2}{sqrt 3}i)$
$frac {1}{sqrt 3}^2 + (-frac {sqrt 2}{sqrt 3})^2 = 1$ so there exists a unique $theta$ ($0 le theta < 2pi$) so that $cos theta = frac 1{sqrt 3}$ and $sin theta = -frac {sqrt 2}{sqrt 3}$.
In that case $tan theta = frac {sin theta}{cos theta} = frac {-frac {sqrt 2}{sqrt 3}}{frac 1{sqrt 3}} = -{sqrt 2}$.
So $theta$ would be $arctan {-sqrt 2}$ with some possible linear transformation to put it in the proper (4th) quadrant. As $arctan$ returns values from $-frac pi 2$ to $ frac pi 2$ (the 4th and 1st quadrant) we are good.
$theta = arctan{-sqrt 2} = -0.304087... pi$. Which so far as I can tell has no rational interpretation.
So $ frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2(cos arctan{-sqrt 2} + sin arctan{-sqrt 2} i) = frac {sqrt 3}2e^{iarctan{-sqrt 2}}$
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add a comment |
$begingroup$
You should write in form $$ z= |z|(cos alpha +isin alpha)$$
The modulus $$|z| = sqrt{{1over 4}+{1over 2}} = {sqrt{3}over2}$$
Since $sin alpha = -{sqrt{2}over sqrt{3}}$ and $cos alpha = -{1over 2sqrt{3}}$ and you will get an $alpha$ by solving this system.
$endgroup$
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
$begingroup$
Yes, you can by drawing $z$ in complex plane, but you will get not an exact value of argument.
$endgroup$
– greedoid
Jan 8 at 19:26
add a comment |
$begingroup$
HINT
You have
$$
re^{it} = rleft(cos t + isin tright),
$$
and note that
$$
r^2 = left(frac12right)^2+left(frac1{sqrt2}right)^2 = frac34,
$$
can you find $t$?
$endgroup$
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
1
$begingroup$
Your first equation is only true if $r=1$. You omitted $r$ on the RHS.
$endgroup$
– MPW
Jan 8 at 19:28
$begingroup$
@MPW corrected, sorry for the typo
$endgroup$
– gt6989b
Jan 8 at 20:58
add a comment |
$begingroup$
In general any complex number $z=x+iy$ can be represented as $re^{itheta}$ where begin{equation} r^2 = |z|^2 = x^2+y^2end{equation} and begin{equation} theta = tan^{-1} Bigl(frac{y}{x}Bigr)end{equation}
Here, begin{equation}
r = sqrt{frac{1}{4}+frac{1}{2}} = frac{sqrt{3}}{2}end{equation}
begin{equation}
theta = tan^{-1}Bigl(frac{-frac{1}{sqrt{2}}}{frac{1}{2}}Bigr) = tan^{-1}(-sqrt{2})
end{equation}
Here, since $ costheta > 0 $ and $ sintheta <0, theta $ is in fourth quadrant. Thus, $ frac{3π}{2}lethetale2π $ or simply,
$theta = -tan^{-1} (sqrt{2})$
Now assume that begin{equation}
tan{x} = u
end{equation}
Thus, begin{equation}
cos = sqrt{frac{1}{1+u^2}}
end{equation}
Thus,
begin{equation}
x = tan^{-1} (u) = cos^{-1}Bigl(sqrt{frac{1}{1+u^2}}Bigr)
end{equation}
Thus,
begin{equation}
cos(tan^{-1} (u)) = frac{1}{sqrt{1+u^2}}
end{equation}
Thus,
begin{equation} cos(tan^{-1}(-sqrt{2})) = frac{1}{sqrt{3}}end{equation}
Thus,begin{equation} sin{theta} =- sqrt{frac{2}{3}}end{equation}
$endgroup$
$begingroup$
Just a slight slip: our point is in the fourth quadrant of the Gaussian plane, so the sine has to be negative.
$endgroup$
– Lubin
Jan 9 at 2:09
$begingroup$
Yes. I am so sorry for that Lubin
$endgroup$
– John Brookfields
Jan 10 at 7:30
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just do it.
$|frac 12 - frac 1{sqrt 2} i| = sqrt {frac 12^2 + frac 12} = frac {sqrt 3} 2$
So $frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2 (frac 1{sqrt 3} - frac {sqrt 2}{sqrt 3}i)$
$frac {1}{sqrt 3}^2 + (-frac {sqrt 2}{sqrt 3})^2 = 1$ so there exists a unique $theta$ ($0 le theta < 2pi$) so that $cos theta = frac 1{sqrt 3}$ and $sin theta = -frac {sqrt 2}{sqrt 3}$.
In that case $tan theta = frac {sin theta}{cos theta} = frac {-frac {sqrt 2}{sqrt 3}}{frac 1{sqrt 3}} = -{sqrt 2}$.
So $theta$ would be $arctan {-sqrt 2}$ with some possible linear transformation to put it in the proper (4th) quadrant. As $arctan$ returns values from $-frac pi 2$ to $ frac pi 2$ (the 4th and 1st quadrant) we are good.
$theta = arctan{-sqrt 2} = -0.304087... pi$. Which so far as I can tell has no rational interpretation.
So $ frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2(cos arctan{-sqrt 2} + sin arctan{-sqrt 2} i) = frac {sqrt 3}2e^{iarctan{-sqrt 2}}$
$endgroup$
add a comment |
$begingroup$
Just do it.
$|frac 12 - frac 1{sqrt 2} i| = sqrt {frac 12^2 + frac 12} = frac {sqrt 3} 2$
So $frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2 (frac 1{sqrt 3} - frac {sqrt 2}{sqrt 3}i)$
$frac {1}{sqrt 3}^2 + (-frac {sqrt 2}{sqrt 3})^2 = 1$ so there exists a unique $theta$ ($0 le theta < 2pi$) so that $cos theta = frac 1{sqrt 3}$ and $sin theta = -frac {sqrt 2}{sqrt 3}$.
In that case $tan theta = frac {sin theta}{cos theta} = frac {-frac {sqrt 2}{sqrt 3}}{frac 1{sqrt 3}} = -{sqrt 2}$.
So $theta$ would be $arctan {-sqrt 2}$ with some possible linear transformation to put it in the proper (4th) quadrant. As $arctan$ returns values from $-frac pi 2$ to $ frac pi 2$ (the 4th and 1st quadrant) we are good.
$theta = arctan{-sqrt 2} = -0.304087... pi$. Which so far as I can tell has no rational interpretation.
So $ frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2(cos arctan{-sqrt 2} + sin arctan{-sqrt 2} i) = frac {sqrt 3}2e^{iarctan{-sqrt 2}}$
$endgroup$
add a comment |
$begingroup$
Just do it.
$|frac 12 - frac 1{sqrt 2} i| = sqrt {frac 12^2 + frac 12} = frac {sqrt 3} 2$
So $frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2 (frac 1{sqrt 3} - frac {sqrt 2}{sqrt 3}i)$
$frac {1}{sqrt 3}^2 + (-frac {sqrt 2}{sqrt 3})^2 = 1$ so there exists a unique $theta$ ($0 le theta < 2pi$) so that $cos theta = frac 1{sqrt 3}$ and $sin theta = -frac {sqrt 2}{sqrt 3}$.
In that case $tan theta = frac {sin theta}{cos theta} = frac {-frac {sqrt 2}{sqrt 3}}{frac 1{sqrt 3}} = -{sqrt 2}$.
So $theta$ would be $arctan {-sqrt 2}$ with some possible linear transformation to put it in the proper (4th) quadrant. As $arctan$ returns values from $-frac pi 2$ to $ frac pi 2$ (the 4th and 1st quadrant) we are good.
$theta = arctan{-sqrt 2} = -0.304087... pi$. Which so far as I can tell has no rational interpretation.
So $ frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2(cos arctan{-sqrt 2} + sin arctan{-sqrt 2} i) = frac {sqrt 3}2e^{iarctan{-sqrt 2}}$
$endgroup$
Just do it.
$|frac 12 - frac 1{sqrt 2} i| = sqrt {frac 12^2 + frac 12} = frac {sqrt 3} 2$
So $frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2 (frac 1{sqrt 3} - frac {sqrt 2}{sqrt 3}i)$
$frac {1}{sqrt 3}^2 + (-frac {sqrt 2}{sqrt 3})^2 = 1$ so there exists a unique $theta$ ($0 le theta < 2pi$) so that $cos theta = frac 1{sqrt 3}$ and $sin theta = -frac {sqrt 2}{sqrt 3}$.
In that case $tan theta = frac {sin theta}{cos theta} = frac {-frac {sqrt 2}{sqrt 3}}{frac 1{sqrt 3}} = -{sqrt 2}$.
So $theta$ would be $arctan {-sqrt 2}$ with some possible linear transformation to put it in the proper (4th) quadrant. As $arctan$ returns values from $-frac pi 2$ to $ frac pi 2$ (the 4th and 1st quadrant) we are good.
$theta = arctan{-sqrt 2} = -0.304087... pi$. Which so far as I can tell has no rational interpretation.
So $ frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2(cos arctan{-sqrt 2} + sin arctan{-sqrt 2} i) = frac {sqrt 3}2e^{iarctan{-sqrt 2}}$
answered Jan 8 at 19:47
fleabloodfleablood
71.2k22686
71.2k22686
add a comment |
add a comment |
$begingroup$
You should write in form $$ z= |z|(cos alpha +isin alpha)$$
The modulus $$|z| = sqrt{{1over 4}+{1over 2}} = {sqrt{3}over2}$$
Since $sin alpha = -{sqrt{2}over sqrt{3}}$ and $cos alpha = -{1over 2sqrt{3}}$ and you will get an $alpha$ by solving this system.
$endgroup$
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
$begingroup$
Yes, you can by drawing $z$ in complex plane, but you will get not an exact value of argument.
$endgroup$
– greedoid
Jan 8 at 19:26
add a comment |
$begingroup$
You should write in form $$ z= |z|(cos alpha +isin alpha)$$
The modulus $$|z| = sqrt{{1over 4}+{1over 2}} = {sqrt{3}over2}$$
Since $sin alpha = -{sqrt{2}over sqrt{3}}$ and $cos alpha = -{1over 2sqrt{3}}$ and you will get an $alpha$ by solving this system.
$endgroup$
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
$begingroup$
Yes, you can by drawing $z$ in complex plane, but you will get not an exact value of argument.
$endgroup$
– greedoid
Jan 8 at 19:26
add a comment |
$begingroup$
You should write in form $$ z= |z|(cos alpha +isin alpha)$$
The modulus $$|z| = sqrt{{1over 4}+{1over 2}} = {sqrt{3}over2}$$
Since $sin alpha = -{sqrt{2}over sqrt{3}}$ and $cos alpha = -{1over 2sqrt{3}}$ and you will get an $alpha$ by solving this system.
$endgroup$
You should write in form $$ z= |z|(cos alpha +isin alpha)$$
The modulus $$|z| = sqrt{{1over 4}+{1over 2}} = {sqrt{3}over2}$$
Since $sin alpha = -{sqrt{2}over sqrt{3}}$ and $cos alpha = -{1over 2sqrt{3}}$ and you will get an $alpha$ by solving this system.
answered Jan 8 at 19:17
greedoidgreedoid
43.7k1155108
43.7k1155108
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
$begingroup$
Yes, you can by drawing $z$ in complex plane, but you will get not an exact value of argument.
$endgroup$
– greedoid
Jan 8 at 19:26
add a comment |
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
$begingroup$
Yes, you can by drawing $z$ in complex plane, but you will get not an exact value of argument.
$endgroup$
– greedoid
Jan 8 at 19:26
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
$begingroup$
Yes, you can by drawing $z$ in complex plane, but you will get not an exact value of argument.
$endgroup$
– greedoid
Jan 8 at 19:26
$begingroup$
Yes, you can by drawing $z$ in complex plane, but you will get not an exact value of argument.
$endgroup$
– greedoid
Jan 8 at 19:26
add a comment |
$begingroup$
HINT
You have
$$
re^{it} = rleft(cos t + isin tright),
$$
and note that
$$
r^2 = left(frac12right)^2+left(frac1{sqrt2}right)^2 = frac34,
$$
can you find $t$?
$endgroup$
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
1
$begingroup$
Your first equation is only true if $r=1$. You omitted $r$ on the RHS.
$endgroup$
– MPW
Jan 8 at 19:28
$begingroup$
@MPW corrected, sorry for the typo
$endgroup$
– gt6989b
Jan 8 at 20:58
add a comment |
$begingroup$
HINT
You have
$$
re^{it} = rleft(cos t + isin tright),
$$
and note that
$$
r^2 = left(frac12right)^2+left(frac1{sqrt2}right)^2 = frac34,
$$
can you find $t$?
$endgroup$
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
1
$begingroup$
Your first equation is only true if $r=1$. You omitted $r$ on the RHS.
$endgroup$
– MPW
Jan 8 at 19:28
$begingroup$
@MPW corrected, sorry for the typo
$endgroup$
– gt6989b
Jan 8 at 20:58
add a comment |
$begingroup$
HINT
You have
$$
re^{it} = rleft(cos t + isin tright),
$$
and note that
$$
r^2 = left(frac12right)^2+left(frac1{sqrt2}right)^2 = frac34,
$$
can you find $t$?
$endgroup$
HINT
You have
$$
re^{it} = rleft(cos t + isin tright),
$$
and note that
$$
r^2 = left(frac12right)^2+left(frac1{sqrt2}right)^2 = frac34,
$$
can you find $t$?
edited Jan 8 at 20:58
answered Jan 8 at 19:18
gt6989bgt6989b
34.3k22455
34.3k22455
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
1
$begingroup$
Your first equation is only true if $r=1$. You omitted $r$ on the RHS.
$endgroup$
– MPW
Jan 8 at 19:28
$begingroup$
@MPW corrected, sorry for the typo
$endgroup$
– gt6989b
Jan 8 at 20:58
add a comment |
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
1
$begingroup$
Your first equation is only true if $r=1$. You omitted $r$ on the RHS.
$endgroup$
– MPW
Jan 8 at 19:28
$begingroup$
@MPW corrected, sorry for the typo
$endgroup$
– gt6989b
Jan 8 at 20:58
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
$begingroup$
I've edited my question, sorry
$endgroup$
– janusz
Jan 8 at 19:24
1
1
$begingroup$
Your first equation is only true if $r=1$. You omitted $r$ on the RHS.
$endgroup$
– MPW
Jan 8 at 19:28
$begingroup$
Your first equation is only true if $r=1$. You omitted $r$ on the RHS.
$endgroup$
– MPW
Jan 8 at 19:28
$begingroup$
@MPW corrected, sorry for the typo
$endgroup$
– gt6989b
Jan 8 at 20:58
$begingroup$
@MPW corrected, sorry for the typo
$endgroup$
– gt6989b
Jan 8 at 20:58
add a comment |
$begingroup$
In general any complex number $z=x+iy$ can be represented as $re^{itheta}$ where begin{equation} r^2 = |z|^2 = x^2+y^2end{equation} and begin{equation} theta = tan^{-1} Bigl(frac{y}{x}Bigr)end{equation}
Here, begin{equation}
r = sqrt{frac{1}{4}+frac{1}{2}} = frac{sqrt{3}}{2}end{equation}
begin{equation}
theta = tan^{-1}Bigl(frac{-frac{1}{sqrt{2}}}{frac{1}{2}}Bigr) = tan^{-1}(-sqrt{2})
end{equation}
Here, since $ costheta > 0 $ and $ sintheta <0, theta $ is in fourth quadrant. Thus, $ frac{3π}{2}lethetale2π $ or simply,
$theta = -tan^{-1} (sqrt{2})$
Now assume that begin{equation}
tan{x} = u
end{equation}
Thus, begin{equation}
cos = sqrt{frac{1}{1+u^2}}
end{equation}
Thus,
begin{equation}
x = tan^{-1} (u) = cos^{-1}Bigl(sqrt{frac{1}{1+u^2}}Bigr)
end{equation}
Thus,
begin{equation}
cos(tan^{-1} (u)) = frac{1}{sqrt{1+u^2}}
end{equation}
Thus,
begin{equation} cos(tan^{-1}(-sqrt{2})) = frac{1}{sqrt{3}}end{equation}
Thus,begin{equation} sin{theta} =- sqrt{frac{2}{3}}end{equation}
$endgroup$
$begingroup$
Just a slight slip: our point is in the fourth quadrant of the Gaussian plane, so the sine has to be negative.
$endgroup$
– Lubin
Jan 9 at 2:09
$begingroup$
Yes. I am so sorry for that Lubin
$endgroup$
– John Brookfields
Jan 10 at 7:30
add a comment |
$begingroup$
In general any complex number $z=x+iy$ can be represented as $re^{itheta}$ where begin{equation} r^2 = |z|^2 = x^2+y^2end{equation} and begin{equation} theta = tan^{-1} Bigl(frac{y}{x}Bigr)end{equation}
Here, begin{equation}
r = sqrt{frac{1}{4}+frac{1}{2}} = frac{sqrt{3}}{2}end{equation}
begin{equation}
theta = tan^{-1}Bigl(frac{-frac{1}{sqrt{2}}}{frac{1}{2}}Bigr) = tan^{-1}(-sqrt{2})
end{equation}
Here, since $ costheta > 0 $ and $ sintheta <0, theta $ is in fourth quadrant. Thus, $ frac{3π}{2}lethetale2π $ or simply,
$theta = -tan^{-1} (sqrt{2})$
Now assume that begin{equation}
tan{x} = u
end{equation}
Thus, begin{equation}
cos = sqrt{frac{1}{1+u^2}}
end{equation}
Thus,
begin{equation}
x = tan^{-1} (u) = cos^{-1}Bigl(sqrt{frac{1}{1+u^2}}Bigr)
end{equation}
Thus,
begin{equation}
cos(tan^{-1} (u)) = frac{1}{sqrt{1+u^2}}
end{equation}
Thus,
begin{equation} cos(tan^{-1}(-sqrt{2})) = frac{1}{sqrt{3}}end{equation}
Thus,begin{equation} sin{theta} =- sqrt{frac{2}{3}}end{equation}
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Just a slight slip: our point is in the fourth quadrant of the Gaussian plane, so the sine has to be negative.
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– Lubin
Jan 9 at 2:09
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Yes. I am so sorry for that Lubin
$endgroup$
– John Brookfields
Jan 10 at 7:30
add a comment |
$begingroup$
In general any complex number $z=x+iy$ can be represented as $re^{itheta}$ where begin{equation} r^2 = |z|^2 = x^2+y^2end{equation} and begin{equation} theta = tan^{-1} Bigl(frac{y}{x}Bigr)end{equation}
Here, begin{equation}
r = sqrt{frac{1}{4}+frac{1}{2}} = frac{sqrt{3}}{2}end{equation}
begin{equation}
theta = tan^{-1}Bigl(frac{-frac{1}{sqrt{2}}}{frac{1}{2}}Bigr) = tan^{-1}(-sqrt{2})
end{equation}
Here, since $ costheta > 0 $ and $ sintheta <0, theta $ is in fourth quadrant. Thus, $ frac{3π}{2}lethetale2π $ or simply,
$theta = -tan^{-1} (sqrt{2})$
Now assume that begin{equation}
tan{x} = u
end{equation}
Thus, begin{equation}
cos = sqrt{frac{1}{1+u^2}}
end{equation}
Thus,
begin{equation}
x = tan^{-1} (u) = cos^{-1}Bigl(sqrt{frac{1}{1+u^2}}Bigr)
end{equation}
Thus,
begin{equation}
cos(tan^{-1} (u)) = frac{1}{sqrt{1+u^2}}
end{equation}
Thus,
begin{equation} cos(tan^{-1}(-sqrt{2})) = frac{1}{sqrt{3}}end{equation}
Thus,begin{equation} sin{theta} =- sqrt{frac{2}{3}}end{equation}
$endgroup$
In general any complex number $z=x+iy$ can be represented as $re^{itheta}$ where begin{equation} r^2 = |z|^2 = x^2+y^2end{equation} and begin{equation} theta = tan^{-1} Bigl(frac{y}{x}Bigr)end{equation}
Here, begin{equation}
r = sqrt{frac{1}{4}+frac{1}{2}} = frac{sqrt{3}}{2}end{equation}
begin{equation}
theta = tan^{-1}Bigl(frac{-frac{1}{sqrt{2}}}{frac{1}{2}}Bigr) = tan^{-1}(-sqrt{2})
end{equation}
Here, since $ costheta > 0 $ and $ sintheta <0, theta $ is in fourth quadrant. Thus, $ frac{3π}{2}lethetale2π $ or simply,
$theta = -tan^{-1} (sqrt{2})$
Now assume that begin{equation}
tan{x} = u
end{equation}
Thus, begin{equation}
cos = sqrt{frac{1}{1+u^2}}
end{equation}
Thus,
begin{equation}
x = tan^{-1} (u) = cos^{-1}Bigl(sqrt{frac{1}{1+u^2}}Bigr)
end{equation}
Thus,
begin{equation}
cos(tan^{-1} (u)) = frac{1}{sqrt{1+u^2}}
end{equation}
Thus,
begin{equation} cos(tan^{-1}(-sqrt{2})) = frac{1}{sqrt{3}}end{equation}
Thus,begin{equation} sin{theta} =- sqrt{frac{2}{3}}end{equation}
edited Jan 10 at 7:37
answered Jan 8 at 19:35
John BrookfieldsJohn Brookfields
113
113
$begingroup$
Just a slight slip: our point is in the fourth quadrant of the Gaussian plane, so the sine has to be negative.
$endgroup$
– Lubin
Jan 9 at 2:09
$begingroup$
Yes. I am so sorry for that Lubin
$endgroup$
– John Brookfields
Jan 10 at 7:30
add a comment |
$begingroup$
Just a slight slip: our point is in the fourth quadrant of the Gaussian plane, so the sine has to be negative.
$endgroup$
– Lubin
Jan 9 at 2:09
$begingroup$
Yes. I am so sorry for that Lubin
$endgroup$
– John Brookfields
Jan 10 at 7:30
$begingroup$
Just a slight slip: our point is in the fourth quadrant of the Gaussian plane, so the sine has to be negative.
$endgroup$
– Lubin
Jan 9 at 2:09
$begingroup$
Just a slight slip: our point is in the fourth quadrant of the Gaussian plane, so the sine has to be negative.
$endgroup$
– Lubin
Jan 9 at 2:09
$begingroup$
Yes. I am so sorry for that Lubin
$endgroup$
– John Brookfields
Jan 10 at 7:30
$begingroup$
Yes. I am so sorry for that Lubin
$endgroup$
– John Brookfields
Jan 10 at 7:30
add a comment |
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Why the $1$? Why not $r(cosalpha+isinalpha)$?
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– Lord Shark the Unknown
Jan 8 at 19:16
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hint: find modulus first then divide real part by modulus to get cosine, divide imaginary part by modulus to get sine
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– Vasya
Jan 8 at 19:16
1
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What's wrong with taking the arctangent? You can divide the result by $pi$ to so if the result is a nice fraction (it isn't). You can draw a picture and do some symilar triangle relationship chasing but I don't think you get any "nice" angles.
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– fleablood
Jan 8 at 19:50
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Our professor told us to do so, but I suppose there must be mistake if such "horrible" angles occur.
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– janusz
Jan 8 at 19:52