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Question: How do you multiply two complex numbers in polar form?...

How do you multiply two complex numbers in polar form?

Explanation

Solution

For this question we need to multiply two complex numbers in polar form. So we will first assume two complex numbers lets say z1{{z}_{1}}, z2{{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 i2=−1{{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 z1{{z}_{1}}, z2{{z}_{2}} are two complex numbers in polar form.
So we can write z1{{z}_{1}}, z2{{z}_{2}} as
z1=a(cos⁡α+isin⁡α){{z}_{1}}=a\left( \cos \alpha +i\sin \alpha \right),
z2=b(cos⁡β+isin⁡β){{z}_{2}}=b\left( \cos \beta +i\sin \beta \right)
Now multiplying the two complex numbers z1{{z}_{1}}, z2{{z}_{2}} in polar form, then we will get
⇒z1.z2=[a(cos⁡α+isin⁡α)][b(cos⁡β+isin⁡β)]\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 aa with constant bb, then we will have
⇒z1.z2=ab[(cos⁡α+isin⁡α)(cos⁡β+isin⁡β)]\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 (a+b)(c+d)=ac+ad+bc+bd\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd. Applying this formula in the above equation, then we will get
⇒z1.z2=ab[cos⁡αcos⁡β+isin⁡αcos⁡β+icos⁡αsin⁡β+i2sin⁡αsin⁡β]\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 ii as common from the middle two terms, then we will have
⇒z1.z2=ab[cos⁡αcos⁡β+i(sin⁡αcos⁡β+cos⁡αsin⁡β)+i2sin⁡αsin⁡β]\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 i2=−1{{i}^{2}}=-1. Substituting this value in the above equation, then we will get
⇒z1.z2=ab[cos⁡αcos⁡β+i(sin⁡αcos⁡β+cos⁡αsin⁡β)−sin⁡αsin⁡β]\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
⇒z1.z2=ab[(cos⁡αcos⁡β−sin⁡αsin⁡β)+i(sin⁡αcos⁡β+cos⁡αsin⁡β)]\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⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y and sin⁡(x+y)=sin⁡xcos⁡y−cos⁡xsin⁡y\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
⇒z1.z2=ab[cos⁡(α+β)+isin⁡(α+β)]\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
⇒z1.z2=abcis(α+β)\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 2cisπ2cis\pi with 3cis2π3cis2\pi then we can write the product of these two as
⇒2cisπ(3cis2π)=6cis(π+2π) ⇒2cisπ(3cis2π)=6cis3π \begin{aligned} & \Rightarrow 2cis\pi \left( 3cis2\pi \right)=6cis\left( \pi +2\pi \right) \\\ & \Rightarrow 2cis\pi \left( 3cis2\pi \right)=6cis3\pi \\\ \end{aligned}