Complex numbers $\left(\frac{1+i}{1-i}\right)^k = 1$ what is $k$?

Question

The smallest possible integer $k$ for which $\left(\frac{1+i}{1-i}\right)^k = 1$ is?

I tried solving this, but my answer doesn't match the given answer. Correct me if I'm wrong at some place

My solution:

\begin{align} \left(\frac{1+i}{1-i}.\frac{1+i}{1+i}\right)^k&= 1\\ \left(\frac{1+2i+i^2}{1-i^2}\right)^k&= 1\\ \left(\frac{1+2i-1}{1-(-1)}\right)^k&= 1\\ \left(\frac{2i}{2}\right)^k&= 1\\ i^k&= 1\\ i^4&= 1\\ \end{align}

EDIT: The question is part of the multiple choice section and the answer is 2. Other options include: 4, 8, 16

Answer 1

Your answer looks good. Although $4$ is not the smallest integer such that $i^k=1$, I'm not quite sure what answer you're supposed to give, since there is no smallest integer $k$ with $i^k=1$ ($k$ is a multiple of $4$, but could be $-4$, $-8$, $-12$, etcetera).

Answer 2

You can also try it by: $$\left(\frac{1+i}{1-i}\right)^k = i^k =e^{i{\frac{k\pi}{2}}}= 1 \Rightarrow k=4,8,12，16，……$$

Answer 3

Since:

$$e^{i \pi}+1=0$$ : Euler’s Identity

And with,

$$e^{k i \pi} = -1$$ when $$k \in odd \mathbb{Z}$$

and,

$$e^{k i \pi} = 1$$ when $$k \in even \mathbb{Z}$$

We can simplify as shown by “Xin Fu”

And so any integer divided by 2 that produces an even number will be equal to 1

Answer 4

The result $k=4$ is correct. The given multiple choice answer $k=2$ is wrong, since \begin{align*} \left(\frac{1+i}{1-i}\right)^2=i^2=-1\ne 1 \end{align*}

Answer 5

$(\frac {1+i}{1-i})^k=1$

$(1+i)^k = (1-i)^k$

$\sum_{j=0}^k {k\choose j}i^j = \sum_{j=0}^k {k\choose j}i^j*(-1)^j$

As $(-1)^{even} = 1$ and $(-1)^{odd} = -1$

$\sum_{j=0;j odd}{k\choose j}i^j =0$

As $i^{4k + 1} = i$ and $i^{4k - 1} = -i$

we need to find the smallest $k$ where $\sum_{h=0}^{4h+1 \le k}{k\choose 4h+1} =\sum_{h=1}^{4h-1 \le k}{k\choose 4h-1}$.

${k \choose 4h -1} = {k\choose k -4h + 1}$ so if $k = 4m$ and $g = m-h$ we will have ${k\choose 4h -1} = {k \choose k - 4h + 1} = {k\choose 4(m-h) + 1} = {k\choose 4g + 1}$ so any $k = 4m$ will be a solution so $k =4$ will be a solution.

It's easy to show directly that $k = 1,2,3$ are not:

$1 + i \ne 1 -i$

$(1+i)^2 =1 + 2i -1 \ne 1 - 2i -1 =(1-i)^2$

$(1+i)^3 = 1 + 3i + 3i^2 + i^3 = -2 + 2i \ne 2-2i = 1 - 3i + 3i^2 -i^3 = (1+i)^3$

However $(1 + i)^4 = 1 + 4i + 6i^2 + 4i^3 + i^4 = 1 - 4i^3 + 6i^2 - 4i + i^4 = 1 - 4i + 6i^2 - 4i^3 + i^4 = (1 - i)^4$.