# 2.1: Algebra with Complex Numbers

• • Contributed by Marcia Levitus

The imaginary unit $$i$$ is defined as the square root of -1: $$i= \sqrt{-1}$$. If $$a$$ and $$b$$ are real numbers, then the number $$c= a+ib$$ is said to be complex. The real number $$a$$ is the real part of the complex number $$c$$, and the real number $$b$$ is its imaginary part. If $$a=0$$, then the number is pure imaginary. All the rules of ordinary arithmetic apply with complex numbers, you just need to remember that $$i^2=-1$$ For example, if $$z_1= 2+3i$$ and $$z_2=1-4i$$:

• $$z_1+ z_2=3-i$$
• $$z_1- z_2=1+7i$$
• $$\dfrac{1}{2} z_1+z_2=2-\dfrac{5}{2} i$$
• $$z_1z_2=(2+3i)(1-4i)=2-12 i^2-5i=14-5i$$ (remember that $$i^2 = -1$$!)
• $$z_1^2=(2+3i)(2+3i)=4+9 i^2+12i=-5+12i$$

In order to divide complex numbers we will introduce the concept of complex conjugate.

Complex Conjugate

The complex conjugate of a complex number is defined as the number that has the same real part and an imaginary part which is the negative of the original number. It is usually denoted with a star: If $$z = x + iy$$, then $$z^∗ = x − iy$$

For example, the complex conjugate of $$2-3i$$ is $$2+3i$$. Notice that the product $$zz^*$$ is always real:

$\label{complexeq:eq1}(x+iy)(x-iy)=x^2-ixy+ixy+y^2=x^2+y^2.$

We’ll use this result in a minute. For now, let’s see how the complex conjugate allows us to divide complex numbers with an example:

Example $$\PageIndex{1}$$: Complex Division

Given $$z_1= 2+3i$$ and $$z_2=1-4i$$ obtain $$z_1/z_2$$

Solution

$\dfrac{z_1}{z_2}=\dfrac{2+3i}{1-4i} \nonumber$

Multiply the numerator and denominator by the complex conjugate of the denominator:

$\dfrac{z_1}{z_2}=\dfrac{2+3i}{1-4i}\dfrac{1+4i}{1+4i} \nonumber$

This “trick” ensures that the denominator is a real number, since $$zz^*$$ is always real. In this case,

\begin{align*} (1-4i)(1+4i) &=1-4i+4i-16i^2\\[4pt] &=17. \end{align*}

The numerator is

\begin{align*} (2+3i)(1+4i) &=2+3i+8i+12i^2 \\[4pt] &=-10+11i \end{align*}

Therefore,

\begin{align*}\dfrac{z_1}{z_2} &=\dfrac{2+3i}{1-4i} \\[4pt] &=\displaystyle{\color{Maroon}-\dfrac{10}{17}+\dfrac{11}{17}i} \end{align*}

Example $$\PageIndex{2}$$

Calculate $$(2-i)^3$$ and express your result in cartesian form ($$a + bi$$)

Solution

\begin{align*} (2-i)^3 &= (2-i)(2-i)(2-i) \\[4pt] (2-i)(2-i) &=4-4i+i^2 \\[4pt] &=4-4i-1 \\[4pt] &=3-4i \\[4pt] (2-i)(2-i)(2-i) &=(3-4i)(2-i) \\[4pt] &=6-3i-8i+4i^2 \\[4pt] &=6-11i+4(-1) \\[4pt] &=\displaystyle{\color{Maroon}2-11i} \end{align*}

The concept of complex conjugate is also useful to calculate the real and imaginary part of a complex number. Given $$z = x+iy$$ and $$z^*=x-iy$$, it is easy to see that $$z+z^*=2x$$ and $$z-z^*=2iy$$. Therefore:

$\label{eq2} Re(z)=\dfrac{z+z^*}{2}$

and

$Im(z)=\dfrac{z-z^*}{2i}$

You may wonder what is so hard about finding the real and imaginary parts of a complex number by visual inspection. It is certainly not a problem if the number is expressed as $$a+bi$$, but it can be more difficult when dealing with more complicated expressions.

algebra with complex numbers