I was solving following multiple choice question (more than one options may be correct) on complex analysis.
Question: The function $f(z)=z^2$ where $0 \leq \arg(z) \leq \pi$ is not
(a) even
(b) odd
(c) The one about which nothing can be said
(d) both even and odd
My attempt since elements of domain on $f$ must satisfies $0 \leq \arg(z) \leq \pi$. So we must consider the numbers in lower half plane say $z=a-bi$ is in lower half plane, so that we have $b>0$ then $-z=-a+bi$ will be in the upper half plane and so that its Principal Argument satisfies $0 \leq \arg(z) \leq \pi$. Hence we can operate $f$ on $-z$. Clearly $f(-z)=(-z)^2 =(-a+bi)^2=(-a+bi)(-a+bi)=a^2-b^2-2abi$
On other hand, $f(z)=z^2=(a-bi)^2=a^2-b^2-2abi$ so we see that $f(-z)=f(z)$ for all $z$ that satisfies $0 \leq \arg(z) \leq \pi$ and hence $f$ will be even function?
But answer key says I was wrong. Please help me.