Question

How do you multiply two complex numbers in polar form?

Answer 1

Hint: For this question we need to multiply two complex numbers in polar form. So we will first assume two complex numbers lets say ${{z}_{1}}$, ${{z}_{2}}$. Now we will multiply both the terms and apply the distribution law of multiplication and the well known formula in the complex numbers which is ${{i}^{2}}=-1$. After applying all those formulas and rules we will simplify the equation to get the required solution.

Complete step-by-step solution:
Let us assume ${{z}_{1}}$, ${{z}_{2}}$ are two complex numbers in polar form.
So we can write ${{z}_{1}}$, ${{z}_{2}}$ as
${{z}_{1}}=a\left( \cos \alpha +i\sin \alpha \right)$,
${{z}_{2}}=b\left( \cos \beta +i\sin \beta \right)$
Now multiplying the two complex numbers ${{z}_{1}}$, ${{z}_{2}}$ in polar form, then we will get
$\Rightarrow {{z}_{1}}.{{z}_{2}}=\left[ a\left( \cos \alpha +i\sin \alpha \right) \right]\left[ b\left( \cos \beta +i\sin \beta \right) \right]$
Multiplying the constant $a$ with constant $b$, then we will have
$\Rightarrow {{z}_{1}}.{{z}_{2}}=ab\left[ \left( \cos \alpha +i\sin \alpha \right)\left( \cos \beta +i\sin \beta \right) \right]$
From the distribution law of multiplication, we can write the value of $\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd$. Applying this formula in the above equation, then we will get
$\Rightarrow {{z}_{1}}.{{z}_{2}}=ab\left[ \cos \alpha \cos \beta +i\sin \alpha \cos \beta +i\cos \alpha \sin \beta +{{i}^{2}}\sin \alpha \sin \beta \right]$
Taking $i$ as common from the middle two terms, then we will have
$\Rightarrow {{z}_{1}}.{{z}_{2}}=ab\left[ \cos \alpha \cos \beta +i\left( \sin \alpha \cos \beta +\cos \alpha \sin \beta \right)+{{i}^{2}}\sin \alpha \sin \beta \right]$
In complex numbers we know the value ${{i}^{2}}=-1$. Substituting this value in the above equation, then we will get
$\Rightarrow {{z}_{1}}.{{z}_{2}}=ab\left[ \cos \alpha \cos \beta +i\left( \sin \alpha \cos \beta +\cos \alpha \sin \beta \right)-\sin \alpha \sin \beta \right]$
Rearranging the terms in the above equation, then the above equation is modified as
$\Rightarrow {{z}_{1}}.{{z}_{2}}=ab\left[ \left( \cos \alpha \cos \beta -\sin \alpha \sin \beta \right)+i\left( \sin \alpha \cos \beta +\cos \alpha \sin \beta \right) \right]$
In trigonometry we have the formulas $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$ and $\sin \left( x+y \right)=\sin x\cos y-\cos x\sin y$. Applying these two formulas in the above equation, then we will get
$\Rightarrow {{z}_{1}}.{{z}_{2}}=ab\left[ \cos \left( \alpha +\beta \right)+i\sin \left( \alpha +\beta \right) \right]$
We can also simply write the above solution as
$\Rightarrow {{z}_{1}}.{{z}_{2}}=ab cis\left( \alpha +\beta \right)$

Note: When we are given to multiply two complex numbers in polar form no need to follow the above procedure, we can simply remember the formula and apply this formula to get the result. For example we need to multiply $2cis\pi $ with $3cis2\pi $ then we can write the product of these two as
$\begin{align}
  & \Rightarrow 2cis\pi \left( 3cis2\pi \right)=6cis\left( \pi +2\pi \right) \\
 & \Rightarrow 2cis\pi \left( 3cis2\pi \right)=6cis3\pi \\
\end{align}$