Complex Numbers

Complex numbers are numbers with a real component and an imaginary component. We can represent these using the rectangular form/notation: $$ \begin{equation}\begin{aligned} a+bi\\ \end{aligned}\end{equation} $$

Where $a$ and $b$ are real numbers. The real component is $a$ and the imaginary component is $bi$.

A perspective on real numbers and imaginary numbers

Consider the real number $5$. We can think of it as being a complex number where the imaginary part is $0i$: $$ \begin{equation}\begin{aligned} 5=5+0i\\ \end{aligned}\end{equation} $$

Now consider an imaginary number $3i$. We can think of it as having a real component of $0$: $$ \begin{equation}\begin{aligned} 3i=0+3i\\ \end{aligned}\end{equation} $$

Thus we can think of real numbers as being complex numbers where the imaginary component is $0i$ and of imaginary numbers as being complex numbers where the real component is $0$.

Equality of complex numbers

If we have two complex numbers and we equate them, it is the same as saying that their real components are equal and their imaginary components are equal.

Example

The complex number $z_1=a+bi$ is equal to $3+2i$. Determine the values of $a$ and $b$.

Solution The equality can be written as: $$ \begin{equation}\begin{aligned} a+bi&=3+2i\\ \end{aligned}\end{equation} $$

This means that the real parts are equal: $$ \begin{equation}\begin{aligned} a&=3\\ \end{aligned}\end{equation} $$

And the imaginary parts are equal: $$ \begin{equation}\begin{aligned} bi&=2i\\ \therefore b&=2\\ \end{aligned}\end{equation} $$

Challenge

Find the unknowns in the following, given the equalities:

• $a+3i=2+bi$
• $3+2yi=x+5i$
• $a+bi=-1$

Argand diagrams

We can represent complex numbers on a diagram where the real part, $Re(z)$, is placed on the $x$ axis and the imaginary part, $Im(z)$, is placed on the $y$ axis. Such a diagram is called an Argand diagram.

Drawing a complex number $a+bi$ on the Argand diagram is the same as plotting $(a,b)$ on the Cartesian plane. The only extra detail is that we draw an arrow from the origin to the point.

The modulus, $|z|$, of a complex number

Given a complex number $z=a+bi$, the modulus (length) of $z$ is $$ \begin{equation}\begin{aligned} |z|=\sqrt{a^2+b^2}\\ \end{aligned}\end{equation} $$

This will always produce a positive value, regardless of if $a$ and $b$ are positive or negative.

Example

Find the modulus of the complex number $z_2=1+3i$.

$$ \begin{equation}\begin{aligned} |z_2|&=\sqrt{{1}^2+{3}^2}\\ &=\sqrt{1+9}\\ &=\sqrt{10} \text{ units}\\ \end{aligned}\end{equation} $$

Practice

Calculate the modulus for the following complex numbers:

• $2+i$
• $-2-4i$
• $-1+5i$
• $1-i$
• $-i$

The argument, $\theta$, of a complex number

This is just the angle between the complex number and the positive real axis. The formula for the argument of a complex number is dependent on the quadrant in which the complex number exists but there is always the same acute angle ($\alpha$) being formed with the real axis ($+ve$ or $-ve$): $$ \begin{equation}\begin{aligned} \alpha&=\tan^{-1}|\frac{Im(z)}{Re(z)}|\\ \end{aligned}\end{equation} $$

For the respective quadrants, the argument will be:

Powers of $i$

We close off this lesson by learning to multiply the various powers of $i$. By looking at the first few powers:

• $i^1=i$
• $i^2=(\sqrt{-1})^2=-1$
• $i^3=i^2\cdot i =-1\cdot i=-i$
• $i^4=(i^2)^2=(-1)^2=1$
• $i^5=i^4\cdot i=1\cdot i=i$
• $i^6=i^4\cdot i^2=1\cdot -1 =-1$

If you notice, a pattern is formed in just these few powers: $$i\rightarrow -1\rightarrow -i\rightarrow 1\rightarrow \text(repeat)$$

Appreciate how every fourth power of $i$ is identical to the previous in that sequence:

• $i=i^5=i^9=…=i$
• $i^2=i^6=i^{10}=…=-1$
• $i^3=i^7=i^{11}=…=-i$
• $i^4=i^8=i^{12}=…=1$

We thus know all of the possible outcomes when multiplying $i$ by itself $n$ times. In particular, for the powers of $i$ with multiples of $4$ as the index, the result is always $1$, the multiplicative identity. This means that if we are given a large power of $i$ to evaluate, we just need to identify the largest multiple of $4$ just smaller than or equal to that large power.

Example

Determine the value of $i^{23}$. Because $23$ is not divisible, we simply count down by $1$ until we find a power that is divisible by $4$:

• $22$ is not divisible
• $21$ is not divisible
• $20$ is divisible (the real MVP)

And we know that $23$ is just $20+3$ so we get: $$ \begin{equation}\begin{aligned} i^{23}&=i^{20+3}\\ &=i^{20}\cdot i^{3}\\ \end{aligned}\end{equation} $$

Now we use laws of indices to write the $20$ as a multiple of $4$: $$ \begin{equation}\begin{aligned} i^{20}\cdot i^{3}&=(i^4)^5\cdot i^3\\ \end{aligned}\end{equation} $$

$i^4$ is always $1$: $$ \begin{equation}\begin{aligned} (i^4)^5\cdot i^3&=(1)^5\cdot i^3\\ &=i^3\\ \end{aligned}\end{equation} $$

That’s it! $i^3$ is quotable as $-i$ by now so we can always write it as that or $-\sqrt{-1}$. Thus $i^{23}$ is just $i^3$ in disguise.

Another example

Simplify $i^{50}$. Because $50$ is not divisible by $4$, we count down:

• $49$ is not divisible
• $48$ is not divisible (Jackpot!)

Thus we rewrite the $50$ as $48+2$: $$ \begin{equation}\begin{aligned} i^{50}&=i^{48+2}\\ &=i^{48}\cdot i^{2}\\ \end{aligned}\end{equation} $$

Rewriting $48$ as a multiple of $4$ and simplifying: $$ \begin{equation}\begin{aligned} i^{48}\cdot i^{2}&=(i^4)^{12}\cdot i^2\\ &=(1)^{12}\cdot i^2\\ &=i^2\\ \end{aligned}\end{equation} $$

Done! We know that $i^2$ is just $-1$, thus this seemly intimidating $i^{50}$ is merely a $-1$ in hiding.