How do you write the complex number $ - 2i$ in polar form?

Hint: We are given a complex number and we have to write it in the polar form. For that, we must know the process of converting a complex number into polar form. Complex numbers are of the form $a \pm ib$ , where $a$ is the real part and $ib$ is the imaginary part of the complex number. $r\cos \theta + ir\sin \theta $ is the polar form of the complex number $a + ib$ where $r = \sqrt {{a^2} + {b^2}} $ , $\cos \theta = \dfrac{a}{r}$ and $\sin \theta = \dfrac{b}{r}$ . Using these formulas, we can convert the complex number into polar form.

Complete step-by-step solution:
The complex number given to us is $ - 2i$ , comparing it to the standard form of the complex number $a + ib$ , we get –
$a = 0$ and $b = - 2$
We know that the polar form of $a + ib$ is $r\cos \theta + if\sin \theta $ -
$
   \Rightarrow r = \sqrt {0 + 4} = 2 \\
   \Rightarrow \cos \theta = \dfrac{0}{2} = 0 \\
   \Rightarrow \sin \theta = \dfrac{{ - 2}}{2} = - 1 \\
 $
We know that
$
 \Rightarrow \cos \dfrac{{3\pi }}{2} = 0 \\
   \Rightarrow \cos \theta = \cos \dfrac{{3\pi }}{2} \\
 $
And
$
 \Rightarrow \sin \dfrac{{3\pi }}{2} = - 1 \\
   \Rightarrow \sin \theta = \sin \dfrac{{3\pi }}{2} \\
 $
Hence, the polar form of $ - 2i$ is $2(\cos \dfrac{{3\pi }}{2} + i\sin \dfrac{{3\pi }}{2})$ .

Note: The complex numbers and real numbers are the two types of numbers. A number that can be shown on the number line is known as a real number, and the numbers that cannot be shown on the number line is known as a complex number. We cannot find the square root of negative numbers, so we have supposed $\sqrt { - 1} $ to be iota $(i)$ , this way $\sqrt { - n} $ is written as $\sqrt n i$ . The complex number $a + ib$ is written as $(r,\theta )$. Thus the polar form of $ - 2i$ is written as $(2,\dfrac{{3\pi }}{2})$ . Converting complex numbers into a polar form is used for their expression on the graph paper, so, with a similar approach, we can convert any complex number into polar form.