Dividing Complex Numbers by Complex Numbers

To get rid of an $i$ in the denominator, all you have to do is multiply the numerator and denominator by $i.$ This cancels out $i$ in the denominator. This method is really helpful when you are trying to convert to rectangular form. Let's look at a simple example to illustrate how to do this.

Consider $\dfrac{1+i}{i}.$ To get rid of the $i$ in the denominator, we must multiply both $1+i,$ the numerator, and $i,$ the denominator, by $i.$ $$\begin{align} \dfrac{1+i}{i} &= \dfrac{1+i}{i} \cdot \dfrac{i}{i}\\ \\ &=\dfrac{(1+i)(i)}{(i)(i)}\\ \\ &=\dfrac{i+i^2}{i^2}\\ \\ &=\dfrac{i+(-1)}{(-1)}, \text{ since $i^2=-1$}\\ \\ &=\dfrac{i-1}{-1}\\ \\ &=1-i \end{align}$$

Now, we have a complex number, $1-i,$ in rectangular form! Let's do one more example.

To put $\dfrac{5+3i}{2i}$ into rectangular form, we need to get rid of the $i$ in the denominator. To do this, we multiply both $5i+3,$ the numerator, and $2i,$ the denominator, by $i.$ $$\begin{align} \dfrac{5+3i}{2i} &= \dfrac{5+3i}{2i}\cdot \dfrac{i}{i}\\ \\ &=\dfrac{(5+3i)(i)}{(2i)(i)}\\ \\ &=\dfrac{5i+3i^2}{2i^2}\\ \\ &=\dfrac{5i+3(-1)}{2(-1)}, \text{ since $i^2=-1$}\\ \\ &=\dfrac{5i-3}{-2}\\ \\ &=\dfrac{3-5i}{2}\\ \\ &=\dfrac{3}{2}+\dfrac{-5}{2}i \end{align}$$

Remember, the big takeaway of this section is that, in order to get rid of an unwanted $i$ in the denominator of a fraction with complex numbers, you just need to multiply the numerator and the denominator by $i.$ It makes it easier to convert to rectangular form!

(Video) $i$ in the Denominator

by mathman1024

Here's a video by mathman1024 showing you how to simplify when there's an $i$ in the denominator.

To simplify an expression with an $i$ in the denominator, you multiply both the top and the bottom by $i$. This will cancel out the $i$'s in the denominator and put the $i$ into the numerator. This is what we want.

The first problem he does with an $i$ in the denominator is $\dfrac{14}{2i}$. $$\frac{14}{2i}\\ \frac{14}{2i}\color{red}{*\frac{i}{i}}\\ \frac{14i}{2i^2}\\ \frac{14i}{2(-1)}\\ \frac{14i}{-2}\\ -7i$$

Try the rest of the problems on your own and see if you get the same answers as him.

1. $\dfrac{-18i}{-4i}$
2. $\dfrac{6+20i}{4}$
3. $\dfrac{35-25i}{5i}$

(Videos) Remove $i$ from the Denominator

by mathman1024

Here's a video by mathman1024 showing you how to remove an $i$ from the denominator. Just like with radicals, the common notation is to not have any of the imaginary units in the denominator. So to remove it we multiply both the top and bottom by $i$. This cancels the $i$ in the denominator and puts it into the numerator.

The first example he looks at is: $\dfrac{12}{6i}$. $$\begin{align} \frac{12}{6i}&=\frac{12}{6i}\color{red}{\cdot\frac{i}{i}}\\ &=\frac{12i}{6i^2}\\ &=\frac{12i}{6\color{red}{(-1)}}\\ &=-2i \end{align}$$

Try the next two on your own and then see if you get the same answers.

1. $\dfrac{-20}{5i}$
2. $\dfrac{32i}{4i}$

Here's another video where he goes over some slightly harder problems. The trick with these is to break them into multiply fractions first.

The first problem he looks at is: $\dfrac{42-28i}{7i}$. Because of the numerator we can't just cancel the $i$'s right away. We first break it into two fractions. $$\begin{align} \frac{42-28i}{7i}&=\frac{42}{7i}-\frac{28i}{7i}\\ &=\frac{42}{7i}\color{red}{\cdot\frac{i}{i}}-\frac{28}{7}\\ &=\frac{42i}{7i^2}-4\\ &=\frac{42i}{-7}-4\\ &=-6i-4 \end{align}$$

Try this next one by yourself and see if you get the same answer.

1. $\dfrac{15+36i}{30i}$

In this video he shows you how to remove the $i$ in the denominator when the denominator has multiple terms. To remove them, we multiply the top and the bottom by the complex conjugate of the denominator.

The first example he looks at is: $\dfrac{3}{2+7i}$. So in the denominator we have $2+7i$. We need to multiply by it's complex conjugate. That would be $2-7i$. $$\begin{align} \frac{3}{2+7i}&=\frac{3}{2+7i}\color{red}{\cdot\frac{2-7i}{2-7i}}\\ &=\frac{3(2-7i)}{(2+7i)(2-7i)}\\ &=\frac{6-21i}{4+49}\\ &=\frac{6-21i}{53}\\ &=\frac{6}{53}-\frac{21}{53}i \end{align}$$

Try solving the last one on your own and then check to see if it's correct.

1. $\dfrac{2+7i}{3-4i}$