How to transform the complex number $\frac{(1+i)^{29}}{1-i}$ to the form $a + bi$?

Question

My problem is I have to transform $\displaystyle \frac{(1+i)^{29}}{1-i}$ on its binomial representation $(a + bi)$.

I was thinking about transforming that into it polar representation and then reverse the process to get the binomial one, but it gets tricky! Any help would be really appreciated as I'm studying this subject for one of my finals.

Answer 1

$$\frac{(1+i)^{29}}{1-i}=\frac{(1+i)^{30}}{(1-i)(1+i)}=\frac{[(1+i)^2]^{15}}2$$

Now $(1+i)^2=\cdots=2i$

Answer 2

Multiply top and bottom by the conjugate of the bottom. The conjugate of $ a + b i$ is $ a - b i$.

Answer 3

There are clever tricks you can use here, but the truth is that even large exponents are never very hard.

To calculate $x^{29}$, calculate $x^2$, $x^4$, $x^8$, $x^{16}$ by repeated squaring ($4$ multiplications) and then $x^{29} = x^{16} \cdot x^8 \cdot x^4 \cdot x$ (another $4$ multiplications).

So with no knowledge of what $x$ is at all, you can already reduce a tiresome $28$ operations to only $8$.