Question

How do you multiply complex numbers in trigonometry?

Answer 1

We first take two complex numbers with their principal arguments. We express them both in their exponential and trigonometric form. We also use the trigonometric formulas like $\left( \cos \alpha \cos \beta -\sin \alpha \sin \beta \right)=\cos \left( \alpha +\beta \right);\left( \sin \alpha \cos \beta +\cos \alpha \sin \beta \right)=\sin \left( \alpha +\beta \right)$.

Complete step by step answer:
We have ${{z}_{1}}$ and ${{z}_{2}}$ as two complex numbers with $\alpha ,\beta$ as their principal arguments. We know that $-\pi \le \alpha ,\beta \le \pi$. This range is for the argument of any complex number. We can express any arbitrary complex number as $z={{e}^{i\theta }}$. Here $\theta$ is the argument.
We also can express it as $z={{e}^{i\theta }}=\cos \theta +i\sin \theta$.
We denote ${{z}_{1}}={{e}^{i\alpha }}$ and ${{z}_{2}}={{e}^{i\beta }}$. We also know that $\arg \left( {{z}_{1}}{{z}_{2}} \right)=\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)$.

Now ${{z}_{1}}{{z}_{2}}={{e}^{i\alpha }}.{{e}^{i\alpha }}={{e}^{i\left( \alpha +\beta \right)}}$. We now express it in trigonometry.
${{z}_{1}}{{z}_{2}}=\left( \cos \alpha +i\sin \alpha \right)\left( \cos \beta +i\sin \beta \right)$.
We use formulas like

\Rightarrow \left( \sin \alpha \cos \beta +\cos \alpha \sin \beta \right)=\sin \left( \alpha +\beta \right) \\\ $$ Therefore, $$\left( \cos \alpha +i\sin \alpha \right)\left( \cos \beta +i\sin \beta \right) =\cos \alpha \cos \beta +i\cos \alpha \sin \beta +i\sin \alpha \cos \beta +{{i}^{2}}\sin \alpha \sin \beta \\\ \Rightarrow \left( \cos \alpha +i\sin \alpha \right)\left( \cos \beta +i\sin \beta \right) =\left( \cos \alpha \cos \beta +{{i}^{2}}\sin \alpha \sin \beta \right)+i\left( \cos \alpha \sin \beta +\sin \alpha \cos \beta \right) \\\ \Rightarrow \left( \cos \alpha +i\sin \alpha \right)\left( \cos \beta +i\sin \beta \right) =\left( \cos \alpha \cos \beta -\sin \alpha \sin \beta \right)+i\left( \cos \alpha \sin \beta +\sin \alpha \cos \beta \right) \\\ \therefore \left( \cos \alpha +i\sin \alpha \right)\left( \cos \beta +i\sin \beta \right) =\cos \left( \alpha +\beta \right)+i\sin \left( \alpha +\beta \right) $$ The multiplication for complex numbers in trigonometry works as the summation of the arguments. **Note:** Principal $\arg \left( {{z}_{1}}{{z}_{2}} \right)$ is given by $\alpha +\beta -2\pi $. The same thing can be done for condition of $\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)<-\pi $ by adding $2\pi $ to the argument if it goes less than $-\pi $ to keep it in the range. The complex form can also be represented as $z={{e}^{i\theta }}=\cos \theta +i\sin \theta $.