HSC Extension 1 and 2 Mathematics/4-Unit/Complex numbers

This topic introduces the complex number system from both an algebraic and a geometric point of view. The underlying idea is the introduction of a number $$i\,$$ such that $$i\,$$ and $$-i\,$$ are the two solutions of the quadratic,
 * $$x^2 + 1 = 0\,$$

Multiples of $$i\,$$ are called pure imaginary numbers, and numbers which can be expressed as $$a + bi\,$$ (where $$a\,$$ and $$b\,$$ are real numbers) are called complex numbers. (This means that all real numbers are also complex numbers.) Algebraic manipulation of complex numbers follows all the standard algebraic rules, with the further rule that
 * $$i^2 = -1\,$$

which allows algebraic simplifications.

Real and Imaginary parts
All complex numbers can be written in the form $$a+ib$$ where $$a$$ and $$b$$ are real numbers. In this form, the real part is $$a$$ and the imaginary part is $$b$$. The real part of a complex number $$z$$ is usually denoted $$\mbox{Re}(z)$$, and the imaginary part $$\mbox{Im}(z)$$.

Geometrical representation of complex numbers as a vector (Modulus-argument form)
It is sometimes useful to represent a complex number as a vector. If $$z = x+iy\,$$, one can plot a point on the number plane at $$(x,y)\,$$. The vector from the origin to the point $$(x,y)\,$$ can also be described in terms of the magnitude, or modulus (i.e., the distance from the origin to the point) &mdash; denoted $$|z|\,$$ &mdash; and the angle or argument (i.e., the angle the line through the origin and the point makes with the $$x$$-axis) &mdash; denoted $$\arg z$$. The modulus is usually called $$r$$, and the argument $$\theta$$. The complex number $$z = x+iy\,$$ can then be expressed in the form $$z = r(\cos\theta+i\sin\theta)\,$$, where $$r = \sqrt{x^2+y^2}$$ and $$\theta = \tan^{-1}\frac{y}{x}$$ (WRONG: see discussion). This is known as modulus-argument form and is often abbreviated to $$z = r\mbox{cis}\theta\,$$.

Algebra on complex numbers
This topic requires you to frequently deal algebraically with complex numbers. In order to do this, you must be familiar with the following (which you can read about on Wikipedia's complex number page):
 * Definition of equality of complex numbers, and equating co-efficients
 * Addition, subtraction, and multiplication of complex numbers
 * The conjugate of a complex number, including its algebraic simplifications. You should try to prove some of these properties, by using the forms $$z=x+iy\,$$ and $$\overline{z}=x-iy$$
 * Dealing with complex fractions algebraically

Negative square roots in the quadratic formula
Complex numbers allow us to use the quadratic formula to solve all quadratics, even where the square root is negative. Consider the following quadratic equation, which has no real solutions:
 * $$x^2 + x + 1 = 0\,$$

Using the quadratic formula:
 * $$\begin{align}

x & = \frac{-b \pm \sqrt{b^2 - 4ac} } {2a} \\ & = \frac{-1 \pm \sqrt{-3} } {2} \end{align}$$ We deal with the negative square root as follows:
 * $$\begin{align}

\sqrt{-3} & = \sqrt{3 \times (-1)} \\ & = \sqrt{3} \times \sqrt{-1} \end{align}$$ However, note that
 * $$\sqrt{-1} \neq i\,$$

but rather
 * $$\sqrt{-1} = \pm i\,$$

so we continue with the quadratic formula:
 * $$\begin{align}

x & = \frac{-1 \pm \sqrt{-3} } {2} \\ & = \frac{-1 \pm (\sqrt{3} \times \sqrt{-1})} {2} \\ & = \frac{-1 \pm (\sqrt{3} \times (\pm i))} {2} \\ & = \frac{-1 \pm \sqrt{3}i} {2} \\ & = \frac{-1}{2} \pm \frac{\sqrt{3}}{2}i \end{align}$$ The last form is preferred, because it completely separates the real and imaginary parts in the equation. Note that normally, the detail shown here is not required; you are expected to skip from the first to the fourth line in the above working.

$$\sqrt{-1}\,$$ is not unique
Although pointing out
 * $$\sqrt{-1} \neq i\,$$

may seem pedantic, it is an important thing to realise and be aware of. In the real number system, $$x=\sqrt{a}$$ is defined to be the positive solution (for $$x$$) to
 * $$x^2 - a = 0\,$$

in the complex number system, there is no useful sense of positive-ness or negative-ness, so the square root cannot be uniquely defined. For all intents and purposes, $$i\,$$ could be defined to be 'the other' root of $$x^2 + 1 = 0$$, and dealing with complex numbers wouldn't change.

For a solid example of why it is wrong to outright say $$\sqrt{-1}=i\,$$, consider:
 * $$1=\sqrt{1}=\sqrt{(-1) \times (-1)}=\sqrt{-1} \times \sqrt{-1} = i^2 = -1$$

The mistake lies in
 * mixing the real number (unique) definition of $$\sqrt{x}$$ with the complex number (non-unique) definition
 * assuming $$\sqrt{-1}=i\,$$. If we didn't assume this, then we could write
 * $$\sqrt{-1} \times \sqrt{-1} = (\pm i) \times (\pm i) = \pm i^2 = \pm 1\,$$
 * which at least contains the right answer.

Complex square roots
You should be able to solve equations like
 * $$z^2 = 1+i\,$$

To do this, we first let
 * $$z=a + ib\,$$

and then
 * $$\begin{align}

(a+ib)^2         & = 1 + i \\ a^2 + 2aib + (ib)^2 & = 1 + i \\ a^2 + 2aib + i^2b^2 & = 1 + i \\ a^2 + 2abi - b^2  & = 1 + i \\ \end{align}$$

In order for two complex numbers to be equal, their complex part and their real part must be equal. Therefore, we can equate the coefficients of i (the complex coefficients) and the real numbers separately. From this, we can solve for a and b:

Real Part: $$ a^2 - b^2 = 1 $$

Imaginary Part: $$ 2ab = 1 $$

Soving:
 * $$\begin{align}

b & = \frac{1}{2a} \\ a^2 - \left ( \frac{1}{2a} \right)^2 & = 1\\ a^4 - a^2 - \frac{1}{4} & = 0 \\ \left( a^2 \right)^2 - \left(a^2\right) - \frac{1}{4} & = 0 \\ a^2 & = \frac{1 \pm \sqrt{2}}{2} \\ \end{align}$$ But $$a$$ is real, so $$a^2$$ is positive.
 * $$\begin{align}

a^2 & = \frac{1 + \sqrt{2}}{2} \\ a  & = \pm \sqrt{\frac{1+\sqrt{2}}{2}} \\ b &= \frac{1}{2a} = \pm \sqrt{\frac{1}{2 + 2\sqrt{2}}} \\ z & = a + ib \\ & = \pm \left( \sqrt{\frac{1+\sqrt{2}}{2}} + \sqrt{\frac{1}{2 + 2\sqrt{2}}}i \right) \end{align}$$

We have two solutions for $$z$$, which makes sense, since quadratic expressions have two roots. We can verify that $$z^2 = 1+i$$ by squaring it (which is left as an exercise for the reader).