Further Maths - Complex Numbers

A complex number is a number with both a real and imaginary part.

Introducing $i$

First, a couple of definitions:

Squaring a number is when you multiply it by itself.
A square root of a number is whatever number squared gives the original number.

\[\sqrt{25} = \pm 5 \sqrt{1} = \pm 1\]

But now, what does $\sqrt{-1}$ equal?

\[\sqrt{-1} = i\]

This definition means that:

\[i \times i = -1 -i \times -i = -1\]

Therefore:

\[(i)^3 = -i (i)^4 = 1 (i)^5 = i\]

As you can see, this sequence repeats every 4 powers. You can find the square roots of other numbers by splitting the square root up and just using algebra:

\[\sqrt{-25} = \sqrt{25 \times -1} = \sqrt{25} \times \sqrt{-1}\]

What is the definition of $i$?

$i$ is the number where:

$i^2 = -1

\[\sqrt{-1} = i\]

How often does the series $i^n$ repeat??

Every 4 terms.

What is $i^3$?

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$-i$

What is $i^{75}$?

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$-i$.

What is the solution to $z^2 + 121 = 0$?

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\[11i\]

Complex Numbers

Complex numbers are numbers involving $i$ and a “real” number. For example:

\[3 + 4i 8 - 2i -6 + 5i\]

They are written in the form $a+bi$.

What is the set of complex numbers called?

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$\mathbb{C}$

What does $\mathbb{C}$ represent?

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The set of all complex numbers.

What is the real part of a complex number called?

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$\Re(z)$

What is the imaginary part of a complex number called?

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$\Im(z)$

Q: What is $(5+i)(3+4i)$?

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\[11 + 23i\]

Q: What is $(6+3i)(7+2i)$?

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\[36 + 33i\]

Conjugates

See Further Maths - Conjugates^A.

The Quadratic Formula

At GCSE, the quadratic formula basically had three outcomes depending on the result of $\sqrt{b^2 - 4ac}$:

A surd, meaning you can’t factorise the quadratic
A square number, meaning that you could factorise.
Zero, meaning there was only one solution.
Negative number under the square root, here be dragons.

But at A-Level, we can solve for negative answers using our new friend $i$. Solutions to quadratic equations for complex numbers always come in pairs or conjugates.

An Example

Conider the cubic:

\[z^3 + 9z^2 + 33z + 25\]

Long story short, we get this:

\[(z+1)(z^2 + 8z + 25)\]

To find the solutions for $z^2 + 8z + 25$, we can use completing the square:

\[z^2 + 8z + 25 = 0 (z + 4)^2 - 16 + 25 = 0 (z + 4)^2 + 9 = 0 (z + 4)^2 = -9 (z + 4)^2 = \pm\sqrt{-9} (z + 4)^2 = \pm3i z = -4 \pm 3i\]

So the solutions are:

\[-4 + 3i -4 - 3i\]

Notice how they’re a conjugate pair. This leaves us with a final “factorised quadratic” of:

\[(z + 1)(z - (-4 + 3i))(z - (-4 - 3i))\]

How can you write the brackets for a quadratic with complex solutions $\alpha$ and $\beta$?

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$(z - \alpha)(z - \beta)$

How can you write the brackets for a quadratic with a complex root of $(a+bi)$?

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$(z - (a+bi))(z - (a-bi))$
$(z - a - bi)(z - a + bi)$
$((z-a)+bi)((z-a)-bi)$ - useful for expanding

What’s a useful way of writing a quadratic with complex solutions for bracket expansion?

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$((z-a)+bi)((z-a)-bi)$
Grouping real terms together

What is interesting about complex roots of a quadratic?

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Complex solutions come in pairs
If a quadratic has one complex root, the other root will be its complex conjugate.

What letters are commonly used for complex solutions of polynomials?

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$\alpha$ (alpha), $\beta$ (beta), $\gamma$ (gamma)

What other solution does a polynomial have if it has one complex root?

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Complex solutions come in pairs
If it has one complex root, another root will be its complex conjugate.

Expanding Brackets with Complex Numbers

There are three main ways:

The long, boring way: Multiplying everything out manually.
The slightly less boring way: Treating the conjugates themselves as a term and keeping them grouped together.
Cool kids only route: Treating something like $(z - (-4 + 3i))$ as $((z + 4) - 3i)$

Dividing Complex Numbers

Consider something like:

\[\frac{5+4i}{2-3i}\]

At first, this might look completely non-sensical. There’s not really a way to think about it in this form that makes intuitive sense. However, something like:

\[\frac{5+4i}{2}\]

Makes complete sense, you can just do $\frac{5}{2} + 2i$ because the $2$ divides both the $5$ and the $4$. So in order to divide a complex number, we convert the denominator of the fraction into a real number by exploiting the fact that a complex number $z$ multiplied by its conjugate $z^{\ast}$ is a real number (see Further Maths - Conjugates^A for a reason why).

In order to solve the $\frac{5+4i}{2-3i}$, we multiply both the top and bottom by the conjugate of the denominator.

\[\frac{(5+4i) \times (2+3i)}{(2-3i) \times (2+3i)} = \frac{-2 + 23i}{13}\]

That’s much more manageable. We can then just divide through to get a final solution of:

\[-\frac{2}{13} + \frac{23}{13}i\]

Q: What is $\frac{3-5i}{1+3i}$ in the form $a+bi$?

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Multiply top and bottom by $1 - 3i$.
$(3-5i)(1-3i) = -12 - 14i$
$(1+3i)(1-3i) = 1^2 + 3^2 = 10$

\[\frac{-12 - 14i}{10} = -\frac{6}{5} - \frac{7}{5}i\]

Q: What is $\frac{3+5i}{6-8i}$ in the form $a+bi$?

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Multiply top and bottom by $6+8i$
$(3+5i)(6+8i) = -22 + 54i$
$(6-8i)(6+8i) = 6^2 + 8^2 = 100$

\[\frac{-22 + 54i}{100} = -\frac{11}{50} + \frac{27}{50}\]

2022-05-22

How can you rotate a complex number by 90 degrees?

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Multiply by $\pm i$.