# Multiplying Complex Numbers

Multiplying Complex Numbers
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Have you ever wondered the outcome of the product of a complex number and imaginary number?

To answer this question, let us understand the basics of real numbers, complex numbers and imaginary numbers.

Natural numbers, rational numbers, decimal numbers and irrational numbers are termed as real numbers.

For example, $$0,1, \dfrac{3}{4}, \pi, 0.235, etc$$.

Imaginary numbers are represented by $$\iota$$. The value of $$i\times i=-1$$ or $$\sqrt{-1}=i$$.

A complex number is a combination of real number and an imaginary number.

For example, $$6.2 + 6i$$

In this mini lesson, we will explore the world of multiplication with complex numbers. We will walk through the answers to the questions like how to multiply two complex numbers, how to multiply real number by a complex numbers, how to multiply pure imaginary number by a complex numbers, and how to square a complex number along with solved examples and interactive questions.

## Lesson Plan

 1 How To Multiply Two Complex Numbers? 2 Important Notes on Multiplying Complex Numbers 3 Solved Examples on Multiplying Complex Numbers 4 Interactive Questions on Multiplying Complex Numbers 5 Challenging Question on Multiplying Complex Numbers

## How To Multiply Two Complex Numbers?

Multiplying complex numbers is similar to multiplying polynomials. We use following polynomial identitiy to solve the multiplication.

$$(a+b)(c+d) = ac + ad + bc + bd$$

For multiplying complex numbers we will use the same polynomial identitiy in the follwoing manner.

For example, consider two complex numbers (4 + 2i) and (1 + 6i). If you multiply them, then you get:

\begin{align} &=(4 + 2i)(1 + 6i)\\ &=4\times1 + 4\times6i + 2i\times1 + 2i\times6i\\\ &= 4 + 24i + 2i + 12i^2 (\rightarrow i^2 = -1)\\\ &=4+26i-12\\\ &=-8+26i \end{align}

Here, -8 is real part, 26i is an imaginary part of a complex number.

## How To Multiply a Real Number With a Complex Number?

There is a simple way to multiply a real number with a complex number. Follow the example below:

\begin{align}&=2(12+6i)\\ &= 2(12)+(2)(6i)\\ &=24 + 12i \end{align}

Here, 24 is real part, 12i is an imaginary part of a complex number. In this case by using the distributive property we learn how to multiply a real number by a complex number.

Let us leran something more tricky and a bit complicated.

## How To Multiply a Pure Imaginary Number By a Complex Number?

Let us consider an example to learn howto multipy pure imaginary number by a complex number.

For example, multiply $$3i(3-7i)$$ and write the resultig number in a complex form i.e. $$a+bi$$

\begin{align} &=3i(3) - 3i(7i)\\ &=9i + 21i^2\end{align}

We have $$9i + 21i^2$$ as our resultant number. But we need to simplify it more to get the resultant number in the form of $$a+bi$$

We know that,

$$i^{2} = -1$$

\begin{align} &=9i - 21i^{2}\\ &=9i - 21(-1)\\ &=9i + 21\end{align}

With the help of commutative property we can write the resultant number as $$21+9i$$

Hence,

$$3i(3-7i)=21+9i$$

## How To Square a Complex Number?

Squaring a complex number is one of the way to multiply a complex number by itself.

\begin{align} &=(1 + 4i)(1 + 4i)\\ &= 1 + 16i + 16i^{2}\\ &= −15 + 16i\end{align}

Hence,

$$(1+4i)^{2}=-15+16i$$

Important Notes
• For multiplication of complex numbers we use:
\begin{align}&=(p_1+q_1i)(p_2+q_2i)\\ &=(p_1p_2-q_1q_2)+i(p_1q_2+p_2q_1)\end{align}
• All real numbers are complex numbers but all complex numbers don't need to be real numbers.
• All imaginary numbers are complex numbers but all complex numbers don't need to be imaginary numbers.
• The conjugate of a complex number $$z=a+ib$$ is $$\bar{z}=a-ib$$
• The magnitude of a complex number $$z=a+ib$$ is $$|z|=\sqrt{a^{2}+b^{2}}$$

## Solved Examples

Let us have a look at solved examples on multiplying complex numbers.

 Example 1

Help Jamie to find the product of the follwoing:
a) $${z_1} = 3 - 2i$$ and $${z_2} = - 4 + 3i$$

b) $${z_1} = 1 + 2i$$ and $${z_2} = 2 + 3i$$

Solution

a) $${z_1} = 3 - 2i$$ and $${z_2} = - 4 + 3i$$

\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}

b) $${z_1} = 1 + 2i$$ and $${z_2} = 2 + 3i$$

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}

 $$\therefore$$ $$(3 - 2i)(- 4 + 3i) = -6 + 17i$$ $$(1+2i)(2+3i) = -4+7i$$
 Example 2

Help Alexa in finding the value of square of $$(-4+6i)^{2}$$.

Solution

Squaring a complex number is one of the way to multiply a complex number by itself.

\begin{align}&=(-4+6i)^{2}\\&=(-4+6i)(-4+6i)\\&=(-4)^2+(-4)(6i)+(6i)(-4)+(6i)^2\\ &=16-24i-24i-36\\&=-20-48i\end{align}.

 $$\therefore$$$$(-4+6i)^2 = -20-48i$$

## Interactive Questions

Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.

Challenging Questions
1. Represent the following complex numbers in a polar form:
a) $$2 + 4i$$
b) $$3 - 6i$$
2. Express the sum, difference, product, and quotient of the following complex numbers as a complex number.
$$z_1=-3+i$$, $$z_2=1+3i$$, and $$z_3=-1-3i$$

## Let's Summarize

The mini-lesson targeted the fascinating concept of multiplying complex numbers. The math journey around multiplying complex numbers starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

## FAQ's on Multiplying Complex Numbers

### Q1. Can you multiply complex numbers?

Yes, we can multiply complex number using the following identity:
\begin{align}&=(p_1+q_1i)(p_2i+q_2i)\\ &=(p_1p_2-q_1q_2)+(p_1q_2+p_2q_1)i\end{align}

For example,

\begin{align} &=(4 + 2i)(1 + 6i)\\ &=4\times1 + 4\times6i + 2i\times1 + 2i\times6i\\\ &= 4 + 24i + 2i + 12i^2 (\rightarrow i^2 = -1)\\\ &=4+26i-12\\\ &=-8+26i \end{align}

Here, -8 is real part, 26i is an imaginary part of a complex number.

### Q2. Is 0 a complex number?

0 is a complex as well as a real number.

### Q3. What is the formula for multiplication of complex numbers?

We can multiply complex number using the following formula:
\begin{align}&=(p_1+q_1i)(p_2i+q_2i)\\ &=(p_1p_2-q_1q_2)+(p_1q_2+p_2q_1)i\end{align}

### Q4. What is the product of two complex numbers?

Product of two complex numbers will be a complex number.
\begin{align}&=(p_1+q_1i)(p_2i+q_2i)\\ &=(p_1p_2-q_1q_2)+(p_1q_2+p_2q_1)i\end{align}

### Q5. How do you multiply complex numbers in polar form?

In polar form, when we multiply a complex number, we need to multiply the magnitudes and add the respective angles.

### Q6. How do you square a complex number?

Squaring a complex number is one of the way to multiply a complex number by itself. For example,

\begin{align} &=(1 + 4i)(1 + 4i)\\ &= 1 + 16i + 16i^{2}\\ &= −15 + 16i\end{align}

Hence,

$$(1+4i)^{2}=-15+16i$$

### Q7. What is i2 equal to?

The value of $$i^{2}$$ or $$i\times i=-1$$ as $$\sqrt{-1}=i$$

### Q8. What is the square root of i in complex numbers?

The value of $$i\times i=-1$$ or $$\sqrt{-1}=i$$

$$\sqrt{i}=\pm(\dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2})$$.

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