Finding the smallest $n$ that satisfies $\left(\frac{1-i}{\sqrt{2}}\right)^n=1$ Maybe I'm just stupid, but right now I'm in a dilemma so please help. I had the question

$$\left(\frac{1-i}{\sqrt{2}}\right)^n=1$$ where I need to find the smallest number for $n$ ($n> 0$) that would satisfy the problem.

My first step was to $$\left[\left(\frac{1-i}{\sqrt{2}}
\right)^2\right]^\frac{n}{2}=1$$ and than $$(-i)^\frac{n}{2}=1$$ after that just ignore the minus because its powered by a even number and so on so forth $$i^\frac{n}{2}=1$$ $$\frac{n}{2}=4$$the  answer is $$n=8$$. But here is where the dilemma came if $$-i=i^3$$ than it would be $$\left(i^3\right)^\frac{n}{2}=1$$ which turns to $$i^\frac{3n}{2}=1$$ and the $$\frac{3n}{2}=4$$. This seems alright in the head but than the answer turns out to be $n=\frac{8}{3}$.
I used latex. Sorry if bad latex i'ts my first time.
If you know the answer please please tell me.
By the way I'm 17 and dumb. So please don't hate
 A: I hope you'll look past the many responses you've been given that presume calculus.  It is simply much easier dealing with complex numbers after Euler's identity, once you get there.
Your question, however, isn't how to work this problem, but that working it two different ways is giving you different results.  When that happens, it is a good idea to narrow down where the error is.
First of all, $n=8/3$ isn't a valid solution for the original problem.  You can confirm that on any calculator sufficiently advanced to deal with complex numbers.  Using $n=8/3$ doesn't give $1$, but rather is close to $-0.5 - 0.866i$.  So you made a mistake somewhere.
Up through the $(i^3)^{\frac{n}{2}} = 1$ the value $n=8/3$ does not work, and the value of $n=8$ does.  But that changes at $i^{\frac{3n}{2}} = 1$.  Now $n=8/3$ works.  What about this operation is broken?  Something about the move from here:
$$
(i^3)^{\frac{n}{2}} = 1
$$
to here:
$$i^{\frac{3n}{2}} = 1$$
is not as equivalent as you thought.
HINT: The rule $\sqrt{xz} = \sqrt{x}\cdot\sqrt{z}$ only holds for real numbers, not complex numbers.  We have to be careful when taking roots of complex numbers.
HINT #2: Normally, to isolate n, we would end up with $n=log(something)$.  You couldn't have known this yet, but logarithms have to be treated with care in complex numbers.  You'll get to this later, but multiplying a complex number by a real number scales it, whereas multiplying by an imaginary number rotates it (and multiplying by a complex number does both).  From that much information, can you infer why a logarithm might be problematic?
A: $Z=((1-i)/\sqrt{2})^n=1 \implies e^{-in\pi/4}=1 \implies (e^{-i\pi})^{n/4}=1 \implies (-1)^{n/4}=1 \implies n=8.$
A: Recall how two complex numbers are multiplied: the length of the product is the product of the length as the product's argument, (https://en.wikipedia.org/wiki/Argument_(complex_analysis), is the sum of the factors' arguments.
Considering $z=(1-i)/\sqrt2$; it has length $1$, any power of $z$ has length $1$ two.  As the argument of $z$ is $-45^\circ$ we'll need the eighth power of $z$ to get $1$.
