1. ## Complex numbers

Explain why $z = (1 + i)^n + (1 - i)^n$ is real for all $n \in N$
Thanks in advance for any help

2. Hello,
Originally Posted by janvdl
Thanks in advance for any help
Note that $(1+i)^2=2i$ and $(1-i)^2=-2i$

Then let n=4k, 4k+1, 4k+2, 4k+3, and use properties of exponents. If you're stuck... well you know what to do

3. you have a simple case of $z + \overline{z}$. remember $\overline{z}^n = \overline{z^n}$

Bobak

4. Originally Posted by bobak
you have a simple case of $z + \overline{z}$. remember $\overline{z}^n = \overline{z^n}$

Bobak
And this is especially clear if you consider the polar form of 1 + i and 1 - i and then raise each to the power of n using de Moivre's Theorem.

5. Finally, you could do this using the binomial theorem, together with the fact that $\left(\begin{array}{c}n \\ k\end{array}\right)= \left(\begin{array}{c}n \\ n-k \end{array}\right)$, and combining all odd powers of i.

6. Originally Posted by HallsofIvy
Finally, you could do this using the binomial theorem, together with the fact that $\left(\begin{array}{c}n \\ i\end{array}\right)= \left(\begin{array}{c}n \\ n-i \end{array}\right)$, and combining all odd powers of i.
... but preferably don't use i as an index in a question about complex numbers.

7. Originally Posted by Opalg
... but preferably don't use i as an index in a question about complex numbers.
lol! Or j .....