Conjugate of the k-th power of a complex number The $k$-th power of a complex number $z$ can be expressed as follows:
$$
z^k=(x+iy)^k=\sum^k_{n=0}\binom{k}{n}x^{k-n}(iy)^n=\sum^k_{n=0}\binom{k}{n}x^{k-n}i^ny^n
$$
Suppose I want to express $\bar{z^k}$, the conjugate of $z^k$, in a similar manner. In other words, for all terms containing $i$, I would like to flip the sign. Is the following expression reasonable and self-evident?
$$
\bar{z^k}=\sum^k_{n=0}\binom{k}{n}x^{k-n}(-i)^ny^n
$$
The motivation behind this is in proving that $\bar{z}^k=\bar{z^k}$ without the use of polar form. 
 A: I would not go with Ragnar's answer. Justifying $\overline{i^n}=(-i)^n$ is nothing different from proving what you were asked to prove: $\overline{z^k}=\bar{z}^k$, except if you want to show $\overline{i^n}=(-i)^n$ by discussing all possibilities of $n$ mod $4$.
I would prove $\overline{zw}=\bar{z}\bar{w}$ first, by writing $z=x_1+y_1i$ and $w=x_2+y_2i$ and computing both sides. Then use induction to show $\overline{z^k}=\bar{z}^k$.
A: That seems right to me. Only think you could remark is that $\bar{i^n}=(-i)^n$, although it is trivial.
A: I'd interpret flipping the sign of $i$ as in essence calculating $\bar{z}^k$, the $k$th power of the conjugate of $z$ since:
$$
\bar{z}^k=[x+(-iy)]^k=\sum^k_{n=0}\binom{k}{n}x^{k-n}(-iy)^n=\sum^k_{n=0}\binom{k}{n}x^{k-n}i^ny^n(-1)^n.
$$
Since the $(-1)^n$ flips the sign of the odd powers of $i$ (which all resolve to $i$ or $-i$, of course), you have your answer, as everything else is the same.  Taking the conjugate of the entire expression does just that.
