A complex number is a number with both a real and imaginary part.
See Also
- [[Further Maths - Polar Form]]S
- [[Further Maths - Argand Diagrams]]S
- [[Further Maths - Loci in the Argand Diagram]]S
- [[Further Maths - Regions in the Argand Diagram]]S
- [[Further Maths - Exponential Form of Complex Numbers]]S
- [[Further Maths - Trig Equations with Complex Numbers]]S
- [[Further Maths - Roots of Complex Numbers]]S
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
- $$\sqrt{-1} = i$
How often does the series $i^n$ repeat??
Every 4 terms.
What is $i^3$??
$-i$
What is $i^{75}$??
$-i$.
What is the solution to $z^2 + 121 = 0$??
$$ 11i $$
Complex Numbers
Complex numbers are numbers involving $i$ and a “real” number. For example:
$$ 3 + 4i 8 - 2i -6 + 5i $$
They are written in the form $a+bi$.
What is the set of complex numbers called??
$\mathbb{C}$
What does $\mathbb{C}$ represent??
The set of all complex numbers.
What is the real part of a complex number called??
$\Re(z)$
What is the imaginary part of a complex number called??
$\Im(z)$
Q: What is $(5+i)(3+4i)$??
$$ 11 + 23i $$
Q: What is $(6+3i)(7+2i)$??
$$ 36 + 33i $$
Conjugates
See [[Further Maths - Conjugates]]S .
The Quadratic Formula
At GCSE, the quadratic formula basically had three outcomes depending on the result of $\sqrt{b^2 - 4ac}$:
- A surd, meaning you can’t factorise the quadratic
- A square number, meaning that you could factorise.
- Zero, meaning there was only one solution.
- Negative number under the square root, here be dragons.
But at A-Level, we can solve for negative answers using our new friend $i$. Solutions to quadratic equations for complex numbers always come in pairs or conjugates.
An Example
Conider the cubic:
$$ 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)) $$
How can you write the brackets for a quadratic with complex solutions $\alpha$ and $\beta$??
$(z - \alpha)(z - \beta)$
How can you write the brackets for a quadratic with a complex root of $(a+bi)$??
- $(z - (a+bi))(z - (a-bi))$
- $(z - a - bi)(z - a + bi)$
- $((z-a)+bi)((z-a)-bi)$ - useful for expanding
What’s a useful way of writing a quadratic with complex solutions for bracket expansion??
- $((z-a)+bi)((z-a)-bi)$
- Grouping real terms together
What is interesting about complex roots of a quadratic??
- Complex solutions come in pairs
- If a quadratic has one complex root, the other root will be its complex conjugate.
What letters are commonly used for complex solutions of polynomials??
$\alpha$ (alpha), $\beta$ (beta), $\gamma$ (gamma)
What other solution does a polynomial have if it has one complex root??
- Complex solutions come in pairs
- If it has one complex root, another root will be its complex conjugate.
Expanding Brackets with Complex Numbers
There are three main ways:
- The long, boring way: Multiplying everything out manually.
- The slightly less boring way: Treating the conjugates themselves as a term and keeping them grouped together.
- Cool kids only route: Treating something like $(z - (-4 + 3i))$ as $((z + 4) - 3i)$
Dividing Complex Numbers
Consider something like:
$$ \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} $$
Makes complete sense, you can just do $\frac{5}{2} + 2i$ because the $2$ divides both the $5$ and the $4$. So in order to divide a complex number, we convert the denominator of the fraction into a real number by exploiting the fact that a complex number $z$ multiplied by its conjugate $z^{\ast}$ is a real number (see [[Further Maths - Conjugates]]S for a reason why).
In order to solve the $\frac{5+4i}{2-3i}$, we multiply both the top and bottom by the conjugate of the denominator.
$$ \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 $$
Q: What is $\frac{3-5i}{1+3i}$ in the form $a+bi$??
- Multiply top and bottom by $1 - 3i$.
- $(3-5i)(1-3i) = -12 - 14i$
- $(1+3i)(1-3i) = 1^2 + 3^2 = 10$
$$ \frac{-12 - 14i}{10} = -\frac{6}{5} - \frac{7}{5}i $$
Q: What is $\frac{3+5i}{6-8i}$ in the form $a+bi$??
- Multiply top and bottom by $6+8i$
- $(3+5i)(6+8i) = -22 + 54i$
- $(6-8i)(6+8i) = 6^2 + 8^2 = 100$
$$ \frac{-22 + 54i}{100} = -\frac{11}{50} + \frac{27}{50} $$
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date: 2020-09-03 18:02
tags:
- '@?further-maths'
- '@?public'
- '@?complex-numbers'
title: Further Maths - Complex Numbers