# Complex Numbers

• Jul 8th 2011, 12:04 AM
Blizzardy
Complex Numbers
Hi guys, need help with a question:

Given that z = 2 + 2i, express z* in the exponential form. Hence find the set of the positive interger n for which (z*)^n is purely imaginary.

z* = 2^(3/2) e^[i(-pi/4)]
(z*)^n = 2^(3n/2) e^[i(-npi/4)]

(i) Given that (z*)^n is purely imaginary,
-npi/4 = pi/2, 3pi/2, 5pi/2, ...
-npi/4 = (2m + 1)(pi/2) where m 'is an element of all intergers, and m > than or = 0'
n = -(4m+2)
(But my ans was wrong)

Anyway, if the question wants to find the set of intergers for n for which (z*)^n is real, can I get the answer simply by letting '-npi/4 = 0, pi, 2pi, 3pi,...' ?

• Jul 8th 2011, 12:51 AM
Sudharaka
Re: Complex Numbers
Quote:

Originally Posted by Blizzardy
Hi guys, need help with a question:

Given that z = 2 + 2i, express z* in the exponential form. Hence find the set of the positive interger n for which (z*)^n is purely imaginary.

z* = 2^(3/2) e^[i(-pi/4)]
(z*)^n = 2^(3n/2) e^[i(-npi/4)]

(i) Given that (z*)^n is purely imaginary,
-npi/4 = pi/2, 3pi/2, 5pi/2, ...
-npi/4 = (2m + 1)(pi/2) where m 'is an element of all intergers, and m > than or = 0'

This is incorrect. It should be, $\displaystyle -\frac{n\pi}{4}=2k\pi\pm\frac{\pi}{2}\mbox{ where }k\in Z$. I have shown how to obtain this below.

n = -(4m+2)
(But my ans was wrong)

Anyway, if the question wants to find the set of intergers for n for which (z*)^n is real, can I get the answer simply by letting '-npi/4 = 0, pi, 2pi, 3pi,...' ?

$\displaystyle \displaystyle\overline{z}^n = 2^{\frac{3n}{2}}e^{i\left(-\frac{n\pi}{4}}\right)}=2^{\frac{3n}{2}}\left(\cos \frac{n\pi}{4}-i\sin\frac{n\pi}{4}\right)$
Since $\displaystyle \displaystyle\overline{z}^n$ is purely imaginary,
$\displaystyle \cos \frac{n\pi}{4}=0\Rightarrow -\frac{n\pi}{4}=2k\pi\pm\frac{\pi}{2}\mbox{ where }k\in Z$