Taylor series expansion of $frac{1}{1+j-z^2}$
0
$begingroup$
I know to manipulate $frac{1}{1+j-z^2}$ to get a result in the form of $frac{1}{1-z}$. However, I have only managed to get to $frac{1}{(-z+1)(-z-1)+j}$. I'm not sure how to manipulate this further
complex-analysis taylor-expansion
edited Jan 11 at 12:10
Bernard
122k741116
asked Jan 11 at 11:51
nastyapplesnastyapples
11
$endgroup$
add a comment |
0
$begingroup$
I know to manipulate $frac{1}{1+j-z^2}$ to get a result in the form of $frac{1}{1-z}$. However, I have only managed to get to $frac{1}{(-z+1)(-z-1)+j}$. I'm not sure how to manipulate this further
complex-analysis taylor-expansion
edited Jan 11 at 12:10
Bernard
122k741116
asked Jan 11 at 11:51
nastyapplesnastyapples
11
$endgroup$
2
$begingroup$
Try the substitution $u = z^2/(1+j)$
$endgroup$
– Damien
Jan 11 at 11:59
$begingroup$
thank you, the substitution worked
$endgroup$
– nastyapples
Jan 11 at 14:25
$begingroup$
You welcome, nice having helped you
$endgroup$
– Damien
Jan 11 at 14:41
add a comment |
0
0
0
$begingroup$
I know to manipulate $frac{1}{1+j-z^2}$ to get a result in the form of $frac{1}{1-z}$. However, I have only managed to get to $frac{1}{(-z+1)(-z-1)+j}$. I'm not sure how to manipulate this further
complex-analysis taylor-expansion
edited Jan 11 at 12:10
Bernard
122k741116
asked Jan 11 at 11:51
nastyapplesnastyapples
11
$endgroup$
I know to manipulate $frac{1}{1+j-z^2}$ to get a result in the form of $frac{1}{1-z}$. However, I have only managed to get to $frac{1}{(-z+1)(-z-1)+j}$. I'm not sure how to manipulate this further
complex-analysis taylor-expansion
complex-analysis taylor-expansion
edited Jan 11 at 12:10
Bernard
122k741116
asked Jan 11 at 11:51
nastyapplesnastyapples
11
edited Jan 11 at 12:10
Bernard
122k741116
asked Jan 11 at 11:51
nastyapplesnastyapples
11
edited Jan 11 at 12:10
Bernard
122k741116
edited Jan 11 at 12:10
Bernard
122k741116
edited Jan 11 at 12:10
Bernard
122k741116
122k741116
asked Jan 11 at 11:51
nastyapplesnastyapples
11
asked Jan 11 at 11:51
nastyapplesnastyapples
11
asked Jan 11 at 11:51
nastyapplesnastyapples
11
11
2
$begingroup$
Try the substitution $u = z^2/(1+j)$
$endgroup$
– Damien
Jan 11 at 11:59
$begingroup$
thank you, the substitution worked
$endgroup$
– nastyapples
Jan 11 at 14:25
$begingroup$
You welcome, nice having helped you
$endgroup$
– Damien
Jan 11 at 14:41
add a comment |
2
$begingroup$
Try the substitution $u = z^2/(1+j)$
$endgroup$
– Damien
Jan 11 at 11:59
$begingroup$
thank you, the substitution worked
$endgroup$
– nastyapples
Jan 11 at 14:25
$begingroup$
You welcome, nice having helped you
$endgroup$
– Damien
Jan 11 at 14:41
2
2
$begingroup$
Try the substitution $u = z^2/(1+j)$
$endgroup$
– Damien
Jan 11 at 11:59
$begingroup$
Try the substitution $u = z^2/(1+j)$
$endgroup$
– Damien
Jan 11 at 11:59
$begingroup$
thank you, the substitution worked
$endgroup$
– nastyapples
Jan 11 at 14:25
$begingroup$
thank you, the substitution worked
$endgroup$
– nastyapples
Jan 11 at 14:25
$begingroup$
You welcome, nice having helped you
$endgroup$
– Damien
Jan 11 at 14:41
$begingroup$
You welcome, nice having helped you
$endgroup$
– Damien
Jan 11 at 14:41
add a comment |
0
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2
$begingroup$
Try the substitution $u = z^2/(1+j)$
$endgroup$
– Damien
Jan 11 at 11:59
$begingroup$
thank you, the substitution worked
$endgroup$
– nastyapples
Jan 11 at 14:25
$begingroup$
You welcome, nice having helped you
$endgroup$
– Damien
Jan 11 at 14:41