System in C

**dhiab** · July 31st 2009, 10:19 PM

Solve in C :

**red_dog** · July 31st 2009, 10:28 PM

Replace $\overline{z}$ in the first equation:

$iz^2=1\Rightarrow z^2=-i=\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}$

Then $z_1=\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}=i\frac{\sqrt{2}}{2}$

$z_2=\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}=\frac{\ sqrt{2}}{2}-i\frac{\sqrt{2}}{2}$

**alexmahone** · July 31st 2009, 10:28 PM

Originally Posted by dhiab

Solve in C :

let $z=a+ib$

a) $\overline{z}z=1$

$(a-ib)(a+ib)=1$

$a^2+b^2=1$ ---------- (1)

b) $iz=\overline{z}$

$i(a+ib)=(a-ib)$

$ai-b=a-ib$

Equating real and imaginary parts, we get

$a=-b$

Substituting this in (1), we get

$2a^2=1$

$a=\pm \frac{1}{\sqrt2}$

$b=\mp \frac{1}{\sqrt2}$

Hence $z=\frac{1}{\sqrt2}-\frac{i}{\sqrt2}$

or $z=-\frac{1}{\sqrt2}+\frac{i}{\sqrt2}$

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