# Further Maths - Complex Numbers > Source: https://ollybritton.com/notes/a-level/further-maths/topics/complex-numbers/ · Updated: 2025-01-08 · Tags: further-maths, complex-numbers > A complex number is a number with both a real and imaginary part. ### See Also - [Further Maths - Polar Form](https://ollybritton.com/notes/a-level/further-maths/topics/polar-form/) - [Further Maths - Argand Diagrams](https://ollybritton.com/notes/a-level/further-maths/topics/argand-diagrams/) - [Further Maths - Loci in the Argand Diagram](https://ollybritton.com/notes/a-level/further-maths/topics/loci-argand-diagram/) - [Further Maths - Regions in the Argand Diagram](https://ollybritton.com/notes/a-level/further-maths/topics/regions-argand-diagram/) - [Further Maths - Exponential Form of Complex Numbers](https://ollybritton.com/notes/a-level/further-maths/topics/exponential-form-of-complex-numbers/) - [Further Maths - Trig Equations with Complex Numbers](https://ollybritton.com/notes/a-level/further-maths/topics/trig-equations-with-complex-numbers/) - [Further Maths - Roots of Complex Numbers](https://ollybritton.com/notes/a-level/further-maths/topics/roots-of-complex-numbers/) ### 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: * $i^2 = -1$ * $$\sqrt{-1} = i$$ ^definition-of-imaginary-unit ##### How often does the series $i^n$ repeat?? Every 4 terms. ^powers-of-i-cycle-length ##### What is $i^3$?? $-i$ ^i-cubed-value ##### What is $i^{75}$?? $-i$. ^i-to-the-75-value ##### What is the solution to $z^2 + 121 = 0$?? $$ 11i $$ ^solution-z-squared-plus-121 ### 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?? $\mathbb{C}$ ^set-of-complex-numbers-symbol ##### What does $\mathbb{C}$ represent?? The set of all complex numbers. ^meaning-of-blackboard-c ##### What is the real part of a complex number called?? $\Re(z)$ ^real-part-notation ##### What is the imaginary part of a complex number called?? $\Im(z)$ ^imaginary-part-notation ##### Q: What is $(5+i)(3+4i)$?? $$ 11 + 23i $$ ^product-5-plus-i-times-3-plus-4i ##### Q: What is $(6+3i)(7+2i)$?? $$ 36 + 33i $$ ^product-6-plus-3i-times-7-plus-2i ### Conjugates See [Further Maths - Conjugates](https://ollybritton.com/notes/a-level/further-maths/topics/conjugates/). ### 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 $\beta$?? $(z - \alpha)(z - \beta)$ ^quadratic-brackets-from-roots ##### 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 ^quadratic-brackets-from-complex-root ##### 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 ^bracket-grouping-for-expansion ##### 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. ^complex-roots-come-in-conjugate-pairs ##### What letters are commonly used for complex solutions of polynomials?? $\alpha$ (alpha), $\beta$ (beta), $\gamma$ (gamma) ^greek-letters-for-polynomial-roots ##### 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. ^conjugate-root-from-single-complex-root ### Expanding Brackets with Complex Numbers There are three main ways: 1. The long, boring way: Multiplying everything out manually. 2. The slightly less boring way: Treating the conjugates themselves as a term and keeping them grouped together. 3. 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](https://ollybritton.com/notes/a-level/further-maths/topics/conjugates/) 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 $1 - 3i$. * $(3-5i)(1-3i) = -12 - 14i$ * $(1+3i)(1-3i) = 1^2 + 3^2 = 10$ $$ \frac{-12 - 14i}{10} = -\frac{6}{5} - \frac{7}{5}i $$ ^divide-3-minus-5i-by-1-plus-3i ##### Q: What is $\frac{3+5i}{6-8i}$ in the form $a+bi$?? * Multiply top and bottom by $6+8i$ * $(3+5i)(6+8i) = -22 + 54i$ * $(6-8i)(6+8i) = 6^2 + 8^2 = 100$ $$ \frac{-22 + 54i}{100} = -\frac{11}{50} + \frac{27}{50}i $$ ^divide-3-plus-5i-by-6-minus-8i ### Rotating complex numbers ##### How can you rotate a complex number by 90 degrees?? Multiply by $\pm i$. ^rotate-complex-number-90-degrees --- Olly Britton — https://ollybritton.com. Machine-readable index: https://ollybritton.com/llms.txt