Using binomial theorem to prove complex conjugate power rule

Question

I am having trouble with the first line of the proof. I am not sure what his happening with the powers on the $b$ term. I thought it would be $(bi)^{n-k}$, but instead it is this:

3. We prove that $\overline{(z^n)}=(\overline z)^n$ (for all positive integers $n$). From the Binomial Theorem, $$\overline{z^n}=\overline{(a+bi)^n}=\overline{\sum_{k=0}^n{n\choose k}a^kb^{n-k}i^{n-k}}$$

Could someone explain this to me?

Answer 1

Starting from

$$\overline{z^n}=\overline{(a+bi)^n}=\overline{\sum_{k=0}^{n}{a^kb^{n-k}i^{n-k}}} \tag{*}$$

and noting that:

• $a,a^k,b,b^{n-k}$ are all real numbers
• the conjugate of a sum is the sum of the conjugates of the terms, i.e. $\overline{x+y}=\overline{x}+\overline{y}$
• the conjugate of a product is the product of the conjugates, i.e. $\overline{x\cdot y}=\overline{x}\cdot\overline{y}$

then ($\ast$) reduces to

$$\overline{z^n}=\sum_{k=0}^{n}{a^kb^{n-k}\overline{i^{n-k}}}$$

Now $\overline{i^{n-k}}=(\overline{i})^{n-k}=(-i)^{n-k}$. So now

$$\overline{z^n}=\sum_{k=0}^{n}{a^kb^{n-k}(-i)^{n-k}}=\sum_{k=0}^{n}{a^k(-bi)^{n-k}}=(a-bi)^n=(\overline{z})^n$$

as required.