Section 5.3 Complex Numbers

Let’s start by providing some background as to where complex numbers arise from. Suppose that we want to solve for the \(x\)-intercepts of \(f(x) = x^2 -2x + 2\text{.}\) We can use the quadratic formula to obtain the \(x\)-intercepts, and when doing so, we obtain the following expression:

Notice that \(\sqrt{-4}\) is not a real number because there are no real numbers whose square is negative. As a result, \(f(x) = x^2 -2x + 2\) has no real \(x\)-intercepts. We can also verify this graphically since the graph of \(f(x) = x^2 - 2x + 2\) never crosses the \(x\)-axis.

Let’s refer back to the motivational example introduced earlier in this section. The value \(\sqrt{-4}\) can be expressed as follows:

Here are more examples of imaginary numbers \(5i\text{,}\) \(\frac{2}{3}i\text{,}\) \(-24i\text{,}\) and \(\sqrt{15}i\text{.}\)

Recall that when solving for the \(x\)-intercepts of \(f(x) = x^2 -2x + 2\text{,}\) we obtained the following values for \(x\) with the quadratic formula:

Using our new established knowledge of imaginary numbers, we find that

which implies that \(x = 1 + i\) or \(x = 1 - i\text{.}\) These \(x\)-values are examples of complex numbers .

When taking the product of two complex numbers, you should always remember that

Finding the product of two pure imaginary numbers is relatively straightforward.

WARNING: The following property is true with real numbers, but it is not true with complex numbers:

Here is a counter-example: let \(a = b = -2\text{.}\) Then,

but

Hence,

(especially when \(a\text{,}\) \(b\) are both negative).

The product of a complex number and its conjugate is a real number. That is,

Now that we have familiarized ourselves with complex numbers, we are ready to make some connections to trigonometry. In the previous subsection, we talked about how to plot a complex number \(z = a+bi\) as a point \((a,b)\) in the complex plane. Moreover, we talked about how the point \((a,b)\) has a distance of \(|z|=\sqrt{a^2+b^2}\) from the origin. We can use our knowledge of polar coordinates to express a complex number in terms of the trigonometric functions \(\sin(\theta)\) and \(\cos(\theta)\text{.}\) Recall that when converting a Cartesian coordinate \((x,y)\) to a polar coordinate \((r,\theta)\text{,}\) we used the following formulas:

The modulus of \(z\) is clearly given by

For convenience, put \(r = |z|\text{.}\) Observe that the ray drawn from the origin to the point \((1,1)\) makes an angle \(\theta\) from the real axis. Then, we can use right triangle trigonometry to write \(a\) and \(b\) as follows:

and

Since \(\tan(\theta) = 1\) and \(a\text{,}\) \(b > 0\text{,}\) the unit circle tells us that \(\theta = \frac{\pi}{4}\text{.}\) This means that

and

Putting things together, the polar form of \(z = 1 + i\) is given by

Observe that the formula for computing the product of two complex numbers in polar form requires us to simply multiply their moduli and add their arguments. Here’s a quick verification of how to obtain the formula for the product using the sum identities for sine and cosine:

We can also use polar form to compute the powers and roots of a complex number with feasibility. For example, if \(z = r(\cos(\theta)+i\sin(\theta))\text{,}\) then using the double-angle formulas we find that

Hence,

De Moivre’s Theorem can also be used for fractional values of \(n\text{,}\) and this allows us to compute the roots of complex numbers. For example, let \(n = \frac{1}{2}\) and \(z = r(\cos(\theta)+i\sin(\theta))\text{.}\) Then, De Moivre’s Theorem gives

Recall that every number (whether it’s real or complex) has two square roots. In the case of complex numbers, we find the other root by adding a multiple of \(2\pi\) to the argument of \(z\text{.}\) So, the polar form of \(z\) can also be given by

Using De Moivre’s Theorem, the second square root of \(z\) is then given by