Further Maths - Complex Numbers
A complex number is a number with both a real and imaginary part.
See Also
- [[Further Maths - Polar Form]]A
- [[Further Maths - Argand Diagrams]]A
- [[Further Maths - Loci in the Argand Diagram]]A
- [[Further Maths - Regions in the Argand Diagram]]A
- [[Further Maths - Exponential Form of Complex Numbers]]A
- [[Further Maths - Trig Equations with Complex Numbers]]A
- [[Further Maths - Roots of Complex Numbers]]A
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.
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
3 + 4i 8 - 2i -6 + 5i
\[They are written in the form $a+bi$. <details class="flashcard"> <summary class="flashcard-front"><p>What is the set of complex numbers called?</p> </summary> <hr> <div class="flashcard-back"> <p>$\mathbb{C}$</p> </div> </details> <details class="flashcard"> <summary class="flashcard-front"><p>What does $\mathbb{C}$ represent?</p> </summary> <hr> <div class="flashcard-back"> <p>The set of all complex numbers.</p> </div> </details> <details class="flashcard"> <summary class="flashcard-front"><p>What is the real part of a complex number called?</p> </summary> <hr> <div class="flashcard-back"> <p>$\Re(z)$</p> </div> </details> <details class="flashcard"> <summary class="flashcard-front"><p>What is the imaginary part of a complex number called?</p> </summary> <hr> <div class="flashcard-back"> <p>$\Im(z)$</p> </div> </details> <details class="flashcard"> <summary class="flashcard-front"><p>Q: What is $(5+i)(3+4i)$?</p> </summary> <hr> <div class="flashcard-back"> \[11 + 23i\] </div> </details> <details class="flashcard"> <summary class="flashcard-front"><p>Q: What is $(6+3i)(7+2i)$?</p> </summary> <hr> <div class="flashcard-back"> \[36 + 33i\] </div> </details>### Conjugates See <a class="wikilink wikilink-notes-a-level notes-a-level-text" href="/notes/a-level/further-maths/topics/conjugates/">[[<span class="notes-a-level-text">Further Maths - Conjugates</span>]]<sup><strong>A</strong></sup></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))
\[<details class="flashcard"> <summary class="flashcard-front"><p>How can you write the brackets for a quadratic with complex solutions $\alpha$ and $\beta$?</p> </summary> <hr> <div class="flashcard-back"> <p>$(z - \alpha)(z - \beta)$</p> </div> </details> <details class="flashcard"> <summary class="flashcard-front"><p>How can you write the brackets for a quadratic with a complex root of $(a+bi)$?</p> </summary> <hr> <div class="flashcard-back"> <ul> <li>$(z - (a+bi))(z - (a-bi))$</li> <li>$(z - a - bi)(z - a + bi)$</li> <li>$((z-a)+bi)((z-a)-bi)$ - useful for expanding</li> </ul> </div> </details> <details class="flashcard"> <summary class="flashcard-front"><p>What’s a useful way of writing a quadratic with complex solutions for bracket expansion?</p> </summary> <hr> <div class="flashcard-back"> <ul> <li>$((z-a)+bi)((z-a)-bi)$</li> <li>Grouping real terms together</li> </ul> </div> </details> <details class="flashcard"> <summary class="flashcard-front"><p>What is interesting about complex roots of a quadratic?</p> </summary> <hr> <div class="flashcard-back"> <ul> <li>Complex solutions come in pairs</li> <li>If a quadratic has one complex root, the other root will be its complex conjugate.</li> </ul> </div> </details> <details class="flashcard"> <summary class="flashcard-front"><p>What letters are commonly used for complex solutions of polynomials?</p> </summary> <hr> <div class="flashcard-back"> <p>$\alpha$ (alpha), $\beta$ (beta), $\gamma$ (gamma)</p> </div> </details> <details class="flashcard"> <summary class="flashcard-front"><p>What other solution does a polynomial have if it has one complex root?</p> </summary> <hr> <div class="flashcard-back"> <ul> <li>Complex solutions come in pairs</li> <li>If it has one complex root, another root will be its complex conjugate.</li> </ul> </div> </details>### 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}
\[Makes complete sense, you can just do $\frac{5}{2} + 2i$ because the $2$ divides both the $5$ and the $4$. So in order to divide a complex number, we convert the denominator of the fraction into a real number by exploiting the fact that a complex number $z$ multiplied by its conjugate $z^{\ast}$ is a real number (see <a class="wikilink wikilink-notes-a-level notes-a-level-text" href="/notes/a-level/further-maths/topics/conjugates/">[[<span class="notes-a-level-text">Further Maths - Conjugates</span>]]<sup><strong>A</strong></sup></a> for a reason why). In order to solve the $\frac{5+4i}{2-3i}$, we multiply both the top and bottom by the conjugate of the denominator.\]\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 $a+bi$?
- Multiply top and bottom by $1 - 3i$.
- $(3-5i)(1-3i) = -12 - 14i$
- $(1+3i)(1-3i) = 1^2 + 3^2 = 10$
Q: What is $\frac{3+5i}{6-8i}$ in the form $a+bi$?
- Multiply top and bottom by $6+8i$
- $(3+5i)(6+8i) = -22 + 54i$
- $(6-8i)(6+8i) = 6^2 + 8^2 = 100$
2022-05-22
How can you rotate a complex number by 90 degrees?
Multiply by $\pm i$.