Question: Find the smallest integer n for which $\displaystyle \left(\frac{1+i}{1-i} \right)^n = 1$
Start by converting to polar coordinates:
$\displaystyle \left(\frac{1+i}{1-i} \right)^n = (\frac{\sqrt 2 \arg 45}{\sqrt 2 \arg -45})^n=\frac{(\sqrt 2)^n \arg (45*n)}{(\sqrt 2)^n \arg (-45*n)}=1 \arg (90*n)$
Now we can see that the solutions are n=...-12,-8,-4,0,4,8,12 ...
Zorro
Can you answer "convert 1+i to polar coordinates". If you can show the notation that you are familiar with then it will be easier to answer your question.
I would say $\displaystyle 1+i=\sqrt 2 \arg {45}$ but you may have been taught to write it in a different way.
$\displaystyle \left( \frac{1 + i}{1 - i} \right)^n$ = $\displaystyle \frac{cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4})}{cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4})}$ = $\displaystyle \frac{(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))}{(cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4}))(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))}$ = $\displaystyle \frac{(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))^2}{cos^2(\frac{n \pi}{4}) + sin^2 (\frac{n \pi}{4})}$ = $\displaystyle \frac{(cos(\frac{n \pi}{2}) + i sin (\frac{n \pi}{2}))}{1}$
Now you want this to equal 1 so the complex part must be zero. That is
$\displaystyle sin (\frac{n \pi}{2})=0$, this occurs if and only if n is even.
Now find the EVEN values of n such that:
$\displaystyle cos (\frac{n \pi}{2})=1$
Hello, zorro!
Find the smallest integer $\displaystyle n$ for which: .$\displaystyle \left(\frac{1+i}{1-i} \right)^n \:=\: 1$
We have: .$\displaystyle \frac{(1+i)^n}{(1-i)^n} \:=\:1 $
Multiply by $\displaystyle (1-i)^n\!:\quad (1+i)^n \:=\:(1-i)^n$
Multiply by $\displaystyle (1 + i)^n\!:\quad (1+i)^n\cdot(1+i)^n \:=\:(1+i)^n(1-i)^n$
. . . . . $\displaystyle (1+i)^{2n} \:=\:\bigg[(1+i)(1-i)\bigg]^n \;=\;\left(1 - i + i - i^2\right)^n $
And we have: .$\displaystyle (1+i)^{2n} \:=\:2^n$
By inspection: .$\displaystyle n \,=\,0$
n=-4000 is an integer less than 0 and $\displaystyle cos(\frac{-4000 \pi}{2})=1$ so -4000 is also a solution. There is no unique smallest solution because for any solution n, n-4 is also a solution.
If your question actually asked for the "smallest positive integer" then the answer would be 4.