Solving exponential complex equation. I need assistance in solving the following equation:

$$ e^{z^2}=1, $$ 

where $z$ is a complex number.. I can't seem to get the answer of $z=\sqrt{k\pi}(1\pm i)$. Thank you in advanced for your help.

We have $\displaystyle e^{z^2}=1=e^{2k\pi i}$ where $k$ is any integer
$\displaystyle\implies z^2=2k\pi i$
$\displaystyle\implies z=\sqrt{2k\pi i}=\sqrt{2k\pi }\sqrt i$
Method $1:$
By observation, $i=\frac{i^2+1+2i}2=\frac{(i+1)^2}2$
$\displaystyle\implies i^{\frac12}=\pm\frac{(1+i)}{\sqrt2}$
Method $2:$
Let $\sqrt{i}=x+iy$ where $x,y$ are real
Squaring we get, $i=(x+iy)^2=x^2-y^2+2xyi$
Equating the real & the imaginary parts we get $x^2-y^2=0$ and $2xy=1$
From the first relation $x=\pm y$
If $x=-y, 1=2xy=2(-y)y\implies y^2=-\frac12$ which is impossible as $y$ is real, $\implies y^2\ge0$
$\implies x=y$ and $1=2xy=2(y)y\implies y^2=\frac12\implies y=\pm\frac1{\sqrt2}$
$\implies \sqrt i=\pm\left(\frac1{\sqrt2}+i\frac1{\sqrt2}\right)=\pm\frac{1+i}{\sqrt2}$
Method $3:$
Let $i=r(\cos\theta+i\sin\theta)=r\cos\theta+ir\sin\theta$ where real $r\ge0$
Equating the real & the imaginary parts we get  $r\cos\theta=0\ \ \ \ $ and $r\sin\theta=1\ \ \ \ (2) $
From $(1),$ either $r=0$ or $\cos\theta=0$
If $r=0,$ from $(2),1=r\sin\theta=0$ which is impossible
$\implies r>0$ and $\cos\theta=0\implies \sin\theta=\pm1$
If $\sin\theta=-1,1=r\sin\theta=-r\iff r=-1$ which is impossible as $r>0$
$\implies \sin\theta=1\implies \theta=2n\pi+\frac\pi2$ where $n$ is any integer
$\implies i=\cos(2n\pi+\frac\pi2)+i(2n\pi+\frac\pi2)=e^{(2n\pi+\frac\pi2)i}$ (using Euler's Formula)
Using  de Moivre's formula,
$\implies i^{\frac12}=e^{\frac{(2n\pi+\frac\pi2)i}2}$ where $n=0,1$
