Would learning Binomial Theorem help with Complex numbers? A high school student asked me whether learning Binomial theorem would help him with complex numbers. At first, I said no but then I realised I was too rusty in the subject to make a judgement. 
What should I tell him? If there is some relevance of Binomial Theorem in complex numbers, to what extent?
Also, specific examples where Binomial Theorem is relevant to complex analysis would be great!
 A: You say "must study" but I don't think it's the right way to look at it. Think of mathematical subjects as tools in a toolkit. As you progress, you gradually accumulate these tools. You never know when a particular tool may come in handy. Before starting on complex number theory (not even analysis), I would say being experienced with all the tools pertaining to basic algebra is important. I think most people would consider binomial expansions (integer powers) to be part of that basic algebra toolkit. So a resounding "yes" from me.
Here are a couple of examples where your student may need to apply binomial theorem when dealing with complex numbers at even an elementary level.
Example 1:
Simplify $(2+3i)^4$, expressing the result in the form $a + bi$.
There is more than one way to handle this problem, and a slightly sophisticated approach would be to transform the number into exponential form before taking the power. However, for such a low exponent, a simple application of binomial theorem is probably easiest. $(2 + 3i) ^4  =2^4 + \binom 4 12^3(3i) + \binom 4 22^2(3i)^2 + \binom 4 32(3i)^3 + (3i)^4$ which can then be easily simplified by evaluating and grouping the real and complex terms together.
Example 2:
As your student delves deeper, he/she will cover de Moivre's theorem, which affords a very easy way to take powers of complex numbers. It's especially useful in the computation of the multiple angle formulae in trigonometry.
For example, suppose you wanted to compute $\cos 3\theta$ in terms of powers of $\cos\theta$.
de Moivre's theorem allows you to immediately state that $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$
Since the real parts on both sides have to be equal, you can use binomial expansion to determine and simplify the real part of the LHS, and then derive $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ without much fuss.
The real power comes when you want to generalise this to any multiple $n$, as in this Wolfram article on Multiple Angle formulae
