# 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}

Complex Numbers
grade 10 | Questions Set 1
Complex Numbers
Complex Numbers