# Division of complex Numbers

Go back to  'Complex-Numbers'

Now, we consider dividing a complex number by another complex number. Take a simple example:

\begin{align}&{z_1} = 4i,\,\,\,{z_2} = 2i\\&\Rightarrow \,\,\,\frac{{{z_1}}}{{{z_2}}} = \frac{{4i}}{{2i}} = 2\end{align}

Now, consider the following two complex numbers:

${z_1} = 4 + 5i,\,\,\,{z_2} = 3i$

We have:

\begin{align}&\frac{{{z_1}}}{{{z_2}}} = \frac{{4 + 5i}}{{3i}} = \frac{4}{{3i}} + \frac{{5i}}{{3i}}\\& = - \frac{4}{3}i + \frac{5}{3} = \frac{5}{3} - \frac{4}{3}i\end{align}

Next, we consider a more advanced case:

${z_1} = 1 + 2i,\,\,\,{z_2} = 3 + 4i$

How do we carry out the division operation and determine $${z_1}/{z_2}$$ in the standard complex form $$a + bi$$? We make the denominator purely real – by multiplying above and below with the conjugate of the denominator:

\begin{align}&\frac{{{z_1}}}{{{z_2}}} = \frac{{1 + 2i}}{{3 + 4i}} = \frac{{\left( {1 + 2i} \right) \times \left( {3 - 4i} \right)}}{{\left( {3 + 4i} \right) \times \left( {3 - 4i} \right)}}\\&= \frac{{1\left( {3 - 4i} \right) + 2i\left( {3 - 4i} \right)}}{{{{\left( 3 \right)}^2} - {{\left( {4i} \right)}^2}}}\\&= \frac{{\left( {3 - 4i} \right) + \left( {6i + 8} \right)}}{{9 + 16}} = \frac{{11 + 2i}}{{25}}\end{align}

Thus,

$\frac{{{z_1}}}{{{z_2}}} = \frac{{11}}{{25}} + \frac{2}{{25}}i$

Example 1: Divide $${z_1} = 4 - 5i$$ by $${z_2} = - 2 + 3i$$.

Solution: We have:

\begin{align}&\frac{{{z_1}}}{{{z_2}}} = \frac{{4 - 5i}}{{ - 2 + 3i}} \times \frac{{ - 2 - 3i}}{{ - 2 - 3i}}\\&= \frac{{ - 8 - 12i + 10i + 15{i^2}}}{{{{\left( { - 2} \right)}^2} + {{\left( 3 \right)}^2}}}\\&= \frac{{ - 23 - 2i}}{{13}} = - \frac{{23}}{{13}} + \left( { - \frac{2}{{13}}} \right)i\end{align}

We note that since dividing two complex numbers gives a complex number, the Complex Set is closed under division as well. However, as should be obvious, division by the complex number 0 is not allowed, as it will lead to a mathematically undefined quantity.

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