Solve $ z = c $ , where $z$ is a complex number and $c$ is an integer using the trigonometric form of z

Question

Knowing that the trigonometric form of any complex number $$ z = r*(cos(\theta)+isin(\theta)) $$ with $ \theta $ a real number.

How do we find $\theta$ for any integer $c$ ?

For example if we have $ z = 1 $ we can write $1$ as $\cos(2k\pi)+i\sin(2k\pi)$ with $k=\overline{0,n}$ with $n$ a natural number , thus makeing $\theta$ be $2k\pi$.

But what if we have numbers like $2$ , $5$ or $8$ , or other complex numbers like $3+4i$ ?

Answer 1

• $2=2\bigl(\cos(\theta)+i\sin(\theta)\bigr)$, with $\theta\in2\pi\mathbb Z$;
• $5=5\bigl(\cos(\theta)+i\sin(\theta)\bigr)$, with $\theta\in2\pi\mathbb Z$;
• $8=8\bigl(\cos(\theta)+i\sin(\theta)\bigr)$, with $\theta\in2\pi\mathbb Z$;
• $3+4i=5\bigl(\cos\theta)+i\sin(\theta)\bigr)$, with $\theta\in\arccos\left(\frac35\right)+2\pi\mathbb Z$.

Answer 2

$$a+bi=\sqrt {a^2+b^2}(\frac {a}{\sqrt {a^2+b^2}}+\frac{b}{\sqrt {a^2+b^2}}i )$$

$$=r(\cos \theta + i \sin \theta )$$

Answer 3

Being $c$ integer is utterly irrelevant. If $$c = a + bi = r(\cos\theta + i\sin\theta),$$ then $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac ba,$$ $$\theta = \cdots$$