Solve: $\displaystyle e^{z^2}=i$
I tried writing it like this: $\displaystyle e^{z^2} = e^{e^{2\log(z)}}=e^{\pi i /2+2\pi i n$
That leaves us with: $\displaystyle e^{2\log(z)} = \pi i /2+2\pi i n$, but that didn't help much.
$\displaystyle e^{z^2} = i$
$\displaystyle e^{(x + iy)^2} = i$
$\displaystyle e^{x^2 + 2ixy +i^2y^2} = i$
$\displaystyle e^{x^2 - y^2 + 2ixy} = i$
$\displaystyle e^{x^2 - y^2}\,\textrm{cis}\,(2xy) = i$
$\displaystyle e^{x^2 - y^2}\cos{(2xy)} + i\,e^{x^2 - y^2}\sin{(2xy)} = 0 + i$.
Equating real and imaginary parts gives:
$\displaystyle e^{x^2 - y^2}\cos{(2xy)} = 0$ and $\displaystyle e^{x^2 - y^2}\sin{(2xy)} = 1$.
Equation 1:
$\displaystyle e^{x^2 - y^2}\cos{(2xy)} = 0$
Since the exponential function is always positive, this means
$\displaystyle \cos{(2xy)} = 0$.
From the Pythagorean Identity:
$\displaystyle \cos^2{(2xy)} + \sin^2{(2xy)} = 1$
$\displaystyle \sin^2{(2xy)} = 1$ since $\displaystyle \cos{(2xy)} = 0$.
Can you go from here?
Well, from where I left off...
$\displaystyle \sin^2{(2xy)} = 1$ since $\displaystyle \cos{(2xy)} = 0$
$\displaystyle \sin{(2xy)} = 1$ or $\displaystyle \sin{(2xy)} = -1$.
Substituting into equation 2:
$\displaystyle e^{x^2 - y^2}\sin{(2xy)} = 1$
Since $\displaystyle e^{x^2 - y^2}$ is always positive, that means the only acceptable solution is $\displaystyle \sin{(2xy)} = 1$.
So $\displaystyle e^{x^2 - y^2} = 1$
$\displaystyle x^2 - y^2 = \ln{1}$
$\displaystyle x^2 - y^2 = 0$
$\displaystyle y^2 = x^2$
$\displaystyle y = \pm x$.
Subtituting back into $\displaystyle \cos{(2xy)} = 0$
$\displaystyle \cos{[2x(\pm x)]} = 0$
$\displaystyle \cos{(\pm 2x^2)} = 0$
$\displaystyle \pm 2x^2 = \frac{\pi}{2} + \pi n$ where $\displaystyle n \in \mathbf{Z}$
$\displaystyle x^2 = \pm\left(\frac{\pi}{4} + \frac{\pi n}{2}\right)$
$\displaystyle x = \pm \sqrt{ \pm(\frac{\pi}{4} + \frac{\pi n}{2} \right ) }$.
Since $\displaystyle y = \pm$ that means $\displaystyle y = \pm \sqrt{\pm \left(\frac{\pi}{4} + \frac{\pi n}{2}\right)}$.
Can you go from here?
That seems to be more solutions than what the textbook says though. This is how I "solved" it, but I'm not sure everything I've done is correct, and I seem to be missing some solutions. I'm especially confused about taking the square root of a complex number, and when you can factor out complex numbers from square roots.
$\displaystyle z^2 = i(\pi /2 + 2 \pi n) \Leftrightarrow z^2 = 2e^{i(\pi /2 + 2 \pi k)}(\pi / 4 + \pi n) \Leftrightarrow$
$\displaystyle z = \pm \sqrt{2}\sqrt{e^{i(\pi /2 + 2 \pi k)}(\pi / 4 + \pi n)} = \pm \sqrt{2}\sqrt{e^{i(\pi /2 + 2 \pi k)}}\sqrt{\pi / 4 + \pi n} = \pm(1+i)\sqrt{\pi / 4 + \pi n}$
I know that this isn't true in general if z and w are complex numbers: $\displaystyle \sqrt{z}\sqrt{w} = \sqrt{zw}$. It's often useful to factor out $\displaystyle -1 = i^2$ from square roots though, so how do you know when you can do it?