Creating a complex number so that its norm equals to 1 I would like to create a complex number c so that its norm is equal to some number a (for the purpose of this question let's assume a = 1) if I already have either its real or imaginary part. I know that:
$$\lVert \mathbf{c} \rVert = \sqrt{\sum_{i=1} ^{n} c_i \overline{c_i}}.$$
In this case:
$$\lVert \mathbf{c} \rVert = \sqrt{(a+bi)(a-bi)} = \sqrt{a^2+abi-abi-b^2i^2} = \sqrt{a^2-b^2\cdot{-1}} = \sqrt{a^2+b^2}.$$
As I said, we already know either a or b, let's say we know that a = 2. How can I find what must b be equal to if the norm must be 1? Some simple equations:
$$ \sqrt{2^2+b^2} = 1 \iff 2^2+b^2 = 1 \iff 4+b^2 = 1 \iff b^2 = 1 - 4 \iff b^2 = -3 \iff b =  \pm\sqrt{-3} \iff $$
$$ \iff b = \pm\sqrt{3} * \sqrt{-1} \iff b = \pm\sqrt{3}i $$
Okay, so based on the above eqations, the imaginary part can be $\pm\sqrt{3}i$. So let's say $c=2+\sqrt{3}i$. But clearly my equations are wrong, because:
$$\sqrt{(2+\sqrt{3}i)(2-\sqrt{3}i)} = \sqrt{7},$$ which does not equal to 1. Please, help me with my confusion and lack of knowledge.
 A: Take any nonzero complex number $c=a+bi$.  Divide by it's length, which is $\sqrt{a^2+b^2}$ to get a complex number of length $1$.
$$z = \frac{c}{||c||}= \frac{a}{\sqrt{a^2+b^2}} + \frac{b}{\sqrt{a^2+b^2}}i.$$
Compute the length of $z$ to see how it works.
A: The Problem
You can't just use any number for $a$ and $b$! If $a$ and $b$ are real numbers, which they are, their square is always positive or zero, which can only be the case if $|a| \leq 1$ and/or $|b| \leq 1$ are less than or equal to 1!
$$
\begin{align*}
a^{2} + b^{2} &= 1 \quad\mid\quad -\left(b^{2}\right)\\
a^{2} &= 1 - b^{2} \quad\mid\quad \sqrt{~~}\\
a &= \sqrt{1 - b^{2}}\\
\end{align*}
$$
If you got an $b > 1$ then will $1 - b^{2}$ get negative and the squareroot of something negativ is not real aka $b$ can't be greater then $1$ but you chose one $b$ or $a$ greater than $1$.
You can also explain it to yourself graphic by imagining the complex number as versor. Of course, the versor is longer (total length) than one [unit] if it is greater than one [unit] in either direction (real part or imaginary part).
Creating complex numbers with the norm $1$
According to the trigonometric Pythagoras there is $\cos(x)^{2} + \sin(x)^2 = 1$ and according to the euler's formula there is $z = e^{x \cdot \mathrm{i}} = \cos(x) + \sin(x) \cdot \mathrm{i}$.
When you combine these you will get $|z| = |e^{x \cdot \mathrm{i}}| = 1$ because $|z| = |e^{x \cdot \mathrm{i}}| =  |\cos(x) + \sin(x) \cdot \mathrm{i}| = \cos(x)^{2} + \sin(x)^2 = 1$.
Aka can chose any $x \in \mathbb{R}$ to get a complex number wich has the norm $1$!
E.G.
$$x = \frac{\pi}{4}$$
$$z = \cos(x) + \sin(x) \cdot \mathrm{i} = \cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4}) \cdot \mathrm{i}$$
$$z = \cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4}) \cdot \mathrm{i} = \frac{1}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}}$$
$$|z| = |\frac{1}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}}| = 1$$
