complex variable question: using exponents I am studing for my final exam, and our prof gave us some questions to do:
1 (a) Express the cubic roots of $-1+i$ in the exponential form. Locate them on a sketch.
(b) Compute $(-1+i)^{16}$ and express your answer in rectangular coordinates. Suggestion: use the exponential form of $-1+i$.
2 (a) Find all values of $i^i$.
(b) Consider the function $f$, such that $f(z)$ is equal to the principal value of $i^z$. Show that $f$ is an entire function.
3) The complex function $\cosh z$ is defined by $\cosh z = \frac{e^{z} + e^{-z}}{2}$.
(a) Prove that $\cosh(z) = \cosh x \cos y+i \sinh x \sin y$, where $x,y \in \mathbb{R}$ and $z = x + iy$.
(b) Assuming that the result of part (a) is already established, verify that the real and imaginary parts of $\cosh z$ satisfy the Cauchy-Riemann equations.
My approach:
1a) $\ln|-1+i| = \sqrt{2}$ and $Arg(-1+i) = 3\pi / 2$, which gives $2^\frac{1}{6} e^{i(\pi/2 + 2/3\pi n)}$.
1b) This turns out to be $2^8 e^{i2\pi n}$, $n \in \mathbb{Z}$.
2a) $i^i = e^{c\log(z)} =e^{i(\pi/2+2\pi n)}$, $n \in \mathbb{Z}$.
2b) I tried to use Cauchy-Riemann Equation to solve this... but had some difficulty... broken the parts to $e^{c \cos x} + ie^{c \sin y}$ and $c=\pi /2+2\pi n$ but CR didn't match...
3a) I think this is straight computation... am I right?
3b) When I expend this, break this into $U(x,y)+iV(x,y)$ and compare $U_x=V_y$ and $U_y=-V_x$?
Any help on this would be appreciated! :) Thank you!
 A: 1) Let $z=-1+i$. Then $|z|=\sqrt{2}$ and $\text{Arg}(z)=\frac{3\pi}{4}$, so $z=|z|e^{i\text{Arg}(z)}=\sqrt{2}e^{i\frac{3\pi}{4}}$ in polar coordinates. Hence $\sqrt[3]{z}=\sqrt[6]{2}e^{i\left(\frac{3\pi}{4}+2k\pi\right)/3}=\sqrt[6]{2}e^{i\left(\frac{\pi}{4}+\frac{2k\pi}{3}\right)}$ for $k=0,1,2$ and $z^{16}=2^8e^{i12\pi}=256$.
2) Use the definition of complex powers:
$$\begin{align*}i^i&=e^{i\ln{i}}=e^{i(\text{Ln}|i|+i\text{Arg}(i)+i2k\pi)}=e^{i(0+i\frac{\pi}{2}+i2k\pi)}=e^{-\frac{\pi}{2}-2k\pi} \text{ for } k\in\mathbb{Z}\\
F(z)&=\text{pv }i^z=\text{pv }e^{z\ln{i}}=e^{z\text{ Ln }{i}}=e^{z(\text{Ln}|i|+i\text{Arg}(i))}=e^{z(0+i\frac{\pi}{2})}=e^{i\frac{\pi z}{2}}=e^{i\frac{\pi}{2}(x+iy)}=e^{-\frac{\pi y}{2}}e^{i\frac{\pi x}{2}}=e^{-\frac{\pi y}{2}}\cos{\frac{\pi x}{2}}+ie^{-\frac{\pi y}{2}}\sin{\frac{\pi x}{2}}=:u(x,y)+iv(x,y)\end{align*}$$
We see that the first partial derivatives of $u$ and $v$ are continuous and satisfy $u_x=v_y$ and $u_y=-v_x$ everywhere in $\mathbb{C}$, so $F(z)$ is an entire function.
3) Recall that $\cosh(x+y)=\cosh{x}\cosh{y}+\sinh{x}\sinh{y}$, $\cosh(iy)=\cos{y}$, and $\sinh(iy)=i\sin{y}$ (these identities can be deduced by writing the functions in exponential forms). We now obtain the desired formula:
$$\cosh{z}=\cosh(x+iy)=\cosh{x}\cosh{iy}+\sinh{x}\sinh{iy}=\cosh{x}\cos{y}+i\sinh{x}\sin{y}=:u(x,y)+iv(x,y)$$
To see that $\cosh{z}:=\frac{e^z+e^{-z}}{2}$ satisfies the Cauchy-Riemann equations, just note that it is a linear combination of two entire functions $e^z$ and $e^{-z}$. Alternatively, show that $u_x=v_y$ and $u_y=-v_x$.
