Assuming $z$ as a complex number,find all solutions to the equation $|i^z|=1$. Assuming $z$ as a complex number,find all solutions to the equation $|i^z|=1$.
My try: here i am confused because for any integer value of z ,it satisfies the equation.
 A: $$i^z = i^{x+iy} = i^x i^{iy} = i^x \cdot (i^i)^y$$
What's $i^i$?  It's
$$i^i = e^{\operatorname{Log}(i^i)} = e^{i\operatorname{Log}i} = e^{i(\ln|i| + i\pi/2)} = e^{i^2\pi/2} = e^{-\pi/2}$$
So then $$|i^z| = |i^x \cdot (i^i)^y| = |i^x \cdot (e^{-\pi/2})^y| = |i^x \cdot e^{-\pi y/2}| = |i^x| \cdot |e^{-\pi y/2}| = 1 \cdot |e^{-\pi y/2}|$$
At this point, we want all values of $y$ such that $|e^{-\pi y/2}| = 1$.  In other words, we want all $y$ such that $e^{-\pi y/2} = \pm 1$.  Well, since $y$ is real, then $e^{-\pi y/2} \ne -1$, and the only time we have $e^{-\pi y/2} = 1$ is when $y=0$.
Thus $x$ can be any real number and $y$ must be zero, which means the solution to $|i^z| = 1$ is $z \in \Bbb R$.
A: Hint:
Use exponential form: $\;i=\mathrm e^{\tfrac{i\pi}2}$, hence
$$i^z=\mathrm e^{\tfrac{z\,i\,\pi}2}$$
If $z=a+ib$, there results
$$\bigl\lvert i^z\bigr\rvert=\mathrm e^{-\tfrac{b\pi}2}.$$
A: Rewrite $i=e^{\frac{i\pi}{2}}$ and $z=a+bi$, so you get
$$ |i^z| =| e^{\frac{i\pi}{2}(a+bi) } |= |e^{\frac{i\pi a}{2} }e^{\frac{-b\pi}{2}}| = |e^{\frac{i\pi a}{2} }| |e^{\frac{-b\pi}{2}}| = |e^{\frac{-b\pi}{2}}| = e^{\frac{-b\pi}{2}} =1=e^0 \to  b = 0$$
so the solution is every real number $a$.
