Multiplication of complex Numbers

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We have learnt how to add and subtract complex numbers. Next, let us understand how to multiply complex numbers. Let’s take a simple example:

\[\begin{align}&{z_1} = 2i,  {z_2} = 3i\\&\Rightarrow \,\,\,{z_1}{z_2} = \left( {2i} \right)\left( {3i} \right) = 6{i^2} =  - 6\end{align}\]

Now, suppose that

\[\begin{align}&{z_1} = 3i,  {z_2} = 2 + i\\&\Rightarrow \,\,\,{z_1}{z_2} \;\;= \left( {3i} \right)\left( {2 + i} \right) = 6i + 3{i^2}\\&\qquad\qquad\;= 6i + 3\left( { - 1} \right) =  - 3 + 6i\end{align}\]

Note how the term 3i distributed over the terms of \({z_2}\). Next, consider

\[{z_1} = 1 + 2i,\,\,\,{z_2} = 2 + 3i\]

To multiply these two numbers, we make use of the distributive law as follows:

\[\begin{align}{z_1}{z_2} &= \left( {1 + 2i} \right)\left( {2 + 3i} \right)\\& = \left( 1 \right)\left( {2 + 3i} \right) + \left( {2i} \right)\left( {2 + 3i} \right)\\&= \left( {2 + 3i} \right) + \left( {4i + 6{i^2}} \right)\\& = \left( {2 + 3i} \right) + \left( {4i - 6} \right)\\& =  - 4 + 7i\end{align}\]

We observe that the product of two complex numbers will also be a complex number - which means that the Complex Set is closed under multiplication.

Example 1: Find the product of \({z_1} = 3 - 2i\) and \({z_2} =  - 4 + 3i\).

Solution: We have:

 \[\begin{align}{z_1}{z_2} &= \left( {3 - 2i} \right)\left( { - 4 + 3i} \right)\\&= 3\left( { - 4 + 3i} \right) - 2i\left( { - 4 + 3i} \right)\\&= \left( { - 12 + 9i} \right) - \left( { - 8i + 6{i^2}} \right)\\&= \left( { - 12 + 9i} \right) - \left( { - 8i - 6} \right)\\&=  - 6 + 17i\end{align}\]

Download practice questions along with solutions for FREE:
Complex Numbers
grade 10 | Questions Set 1
Complex Numbers
grade 10 | Answers Set 1
Complex Numbers
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Complex Numbers
grade 10 | Questions Set 2
Download practice questions along with solutions for FREE:
Complex Numbers
grade 10 | Questions Set 1
Complex Numbers
grade 10 | Answers Set 1
Complex Numbers
grade 10 | Answers Set 2
Complex Numbers
grade 10 | Questions Set 2
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