Calculating a complex sum

Question

I am to solve the following sum: $$\sum_{n=0}^{\infty} \frac{(1+i)^n}{(1+2i)^{2n}}$$ Should I divide the sum on real and imaginary parts (in similar way as limits of sequences are calculated)? If yes how can I do it? The expression is quite complicated.

Answer 1

The series is just

$$\sum_{n = 0}^{\infty} \left(\frac{1 + i}{(1 + 2i)^2}\right)^{n} = \frac 1 {1 - \frac{1 + i}{(1 + 2i)^2}} = \frac{(1 + 2i)^2}{(1 + 2i)^2 - (1 + i)^2}$$

since the modulus of the common ratio is clearly less than $1$. If you want real and imaginary parts, now it's a simple matter of multiplying by conjugates.

Answer 2

HINT

Note that

$$\sum_{n=0}^{\infty} \frac{(1+i)^n}{(1+2i)^{2n}}=\sum_{n=0}^{\infty} \left(\frac{1+i}{(1+2i)^{2}}\right)^n$$