# complex number help

• Dec 10th 2008, 06:05 AM
complex number help
The complex number $z=\frac{1+i}{3-2i}$.Write down the argument for $z^2$ and find the exact value of |z^2| .

My working :

$z=\frac{1}{13}+\frac{5}{13}i$
$z^2=(\frac{1}{13}+\frac{5}{13}i)^2$
$z^2=\frac{-24+10i}{169}$

Up to here , did i make any mistake . My final answer is wrong so i am wondering where my mistake is . Can someone pls check for me ? Thanks .
• Dec 10th 2008, 01:53 PM
Soroban

Quote:

Given: . $z\:=\:\frac{1+i}{3-2i}$.

Find the argument for $z^2$ and find the exact value of $|z^2| .$

My working:

$z\:=\:\frac{1}{13}+\frac{5}{13}i$

$z^2\:=\:\left(\frac{1}{13}+\frac{5}{13}i\right)^2 \;=\;\frac{-24+10i}{169}$

Up to here, did i make any mistakes? .
. . . no

My final answer is wrong. .
You could have shown it to us.

We have: . $z^2 \;=\;-\frac{24}{169} + \frac{10}{169}i$

The argument $\theta$ is given by: . $\tan\theta \;=\;\frac{\frac{10}{169}}{-\frac{24}{169}} \;=\;-\frac{5}{12}$

Hence: . $\theta \;=\;\tan^{-1}\left(\text{-}\tfrac{5}{12}\right) \;\approx\; 2.7468\text{ radians} \;\approx\;157.38^o$

The magnitude of $z^2$ is: . $|z^2| \;=\;\sqrt{\left(\frac{\text{-}24}{169}\right)^2 + \left(\frac{10}{169}\right)^2}\;= \;\sqrt{\frac{676}{169^2}}$

Therefore: . $|z^2| \;=\;\frac{26}{169} \;=\;\frac{2}{13}$