Let $z$ be any complex number. Let $D =\{z: \Re(z) > |\Im(z)|\}$ . Let $f_n(z) = \operatorname{Log} z^n$ where $n = 1,2,3,4$. Then what is $f(D)$
My attempt : I found that $D = \left\{z : \operatorname{Arg}(z) < \frac{\pi}{4}\right\}$ . I wants the solution of the form $f_n (D) = \left\{z : |\Im (z)| < \frac{n\pi}{4}\right\}$ but I don't know how to connect $D = \{z :\operatorname{Arg}(z)<\frac{\pi}{4}\}$ with required result. Please help me.
Here is the proper explanation you requested.
So first, we want $\Re(z)>|\Im(z)|$.The set of $z$ should be $\{z:|\operatorname{Arg}(z)|<\frac{\pi}{4}\}$. Now let $z=re^{i\theta}$, $r\in\mathbb{R}$, $|\theta|<\frac{\pi}{4}$. We then have $$\operatorname{Log}(z)=\ln(r)+i\operatorname{Arg}(z)$$ Remember that $r\in\mathbb{R}$ can take any value, so the image of the function $f_1$ is the set: $$f_1(D)=\left\{x+iy:x\in\mathbb{R}, |y|<\frac{\pi}{4}\right\}$$ Finally, if we consider the cases where $n=2,3,4$, the only thing changes is the value of $|\operatorname{Arg}(z^n)|=n|\operatorname{Arg}(z)|$ ($0<n<5$ so the range of this principal argument works fine).
Therefore, what we got is $$\operatorname{Log}(z^n)=\ln(r^n)+in\operatorname{Arg}(z)$$ Because $r>0$, $r^n$ still has the range of the whole $\mathbb{R}$, therefore $$f_n(D)=\left\{x+iy:x\in\mathbb{R}, |y|<\frac{n\pi}{4}\right\}$$