# 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.

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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 $n=0,1$

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I see. It's annoying that they'd want me to do that extra step to come to the final answer. Shouldn't $z=\pm \sqrt{2k\pi i}$ suffice? – Gustavo Montano Sep 1 '13 at 4:50
@eXtremiity, whenever we find something like $\sqrt{a+ib},$ we need to extract the roots – lab bhattacharjee Sep 1 '13 at 5:32