# Complex numbers from polar to binomic form

How shall we proceed if we want to determine the binomial form of a complex number expressed in the polar form?

Let's see the procedure:

Given now a complex number $$z$$ in polar form $$z=|z|_{\alpha}$$, if we want to find the binomial form we only have to determine $$a$$ and $$b$$, where:

• $$a$$ is the real part and it is: $$a=|z|\cos(\alpha)$$
• $$b$$ the complex part and it is: $$b=|z|\sin(\alpha)$$

For example, if we have the complex number in polar form: $$6_{225^{\circ}}$$.

We can determine the real part of its binomial form by: $$a=6\cos(225^{\circ})=-3\sqrt{2}$$

And the complex part by: $$b=6\sin(225^{\circ})=-3\sqrt{2}$$

Thus we will write it as $$a+ib$$ or using the example: $$-3\sqrt{2}-3\sqrt{2}$$ (that is a binomial form).

We can say in general terms that in order to translate a complex number in polar form into the binomial form, we only have to use the following formula:

$$z_{\alpha}=|z|(\cos\alpha+\sin\alpha \cdot i)$$\$