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.

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.
