Complex Numbers

z = 1 + i/2, 1 - i/2

z^2 = 2i + 2.

let Z = a+bi

Z^2= a^2-b^2+2abi

$a^2-b^2=2 \qquad (1)\\2ab=2\qquad (2)\\ ab=1\\ b=1/a\\ a^2-\frac{1}{a^2}=2\\ a^4-2a^2-1=0\\ let\;\;m=a^2\\ m^2-2m-1=0\\ ...\\ m=1+\sqrt2\qquad or \qquad m=1-\sqrt2\\ \text{but m is positive so}\\ m=1+\sqrt2\\ a=\pm\sqrt{1+\sqrt2}\\ b=\pm\frac{1}{\sqrt{1+\sqrt2}}$

$a=\pm\sqrt{1+\sqrt2}\\ b=\pm\frac{1}{\sqrt{1+\sqrt2}}\\ b=\pm\frac{1}{\sqrt{1+\sqrt2}}\frac{\sqrt{1+\sqrt2}}{\sqrt{1+\sqrt2}}\\ b=\pm\frac{\sqrt{1+\sqrt2}}{1+\sqrt2}\\ b=\pm\frac{\sqrt{1+\sqrt2}}{1+\sqrt2}\times\frac{1-\sqrt2}{1-\sqrt2}\\ b=\pm\frac{\sqrt{1+\sqrt2}}{1-2}\times\frac{\sqrt{1-\sqrt2}\sqrt{1-\sqrt2}}{1}\\ b=\pm\frac{\sqrt{1-2}}{1-2}\times\frac{\sqrt{1-\sqrt2}}{1}\\ b=\pm\frac{i}{-1}\times \frac{i\sqrt{\sqrt2-1}}{1}\\ b=\pm \sqrt{\sqrt2-1}\\ $

You seriously need to check that

PLUS I expect it should ahave been done by letting     $Z=Ae^{i\theta}$

LaTex

a^2-b^2=2 \qquad (1)\\2ab=2\qquad (2)\\
ab=1\\
b=1/a\\
a^2-\frac{1}{a^2}=2\\
a^4-2a^2-1=0\\
let\;\;m=a^2\\
m^2-2m-1=0\\
...\\
m=1+\sqrt2\qquad or \qquad m=1-\sqrt2\\
\text{but m is positive so}\\
m=1+\sqrt2\\
a=\pm\sqrt{1+\sqrt2}\\
b=\pm\frac{1}{\sqrt{1+\sqrt2}}

a=\pm\sqrt{1+\sqrt2}\\
b=\pm\frac{1}{\sqrt{1+\sqrt2}}\\
b=\pm\frac{1}{\sqrt{1+\sqrt2}}\frac{\sqrt{1+\sqrt2}}{\sqrt{1+\sqrt2}}\\
b=\pm\frac{\sqrt{1+\sqrt2}}{1+\sqrt2}\\
b=\pm\frac{\sqrt{1+\sqrt2}}{1+\sqrt2}\times\frac{1-\sqrt2}{1-\sqrt2}\\
b=\pm\frac{\sqrt{1+\sqrt2}}{1-2}\times\frac{\sqrt{1-\sqrt2}\sqrt{1-\sqrt2}}{1}\\
b=\pm\frac{\sqrt{1-2}}{1-2}\times\frac{\sqrt{1-\sqrt2}}{1}\\
b=\pm\frac{i}{-1}\times \frac{i\sqrt{\sqrt2-1}}{1}\\
b=\pm \sqrt{\sqrt2-1}\\
 

Here's an alternative approach: