# An Introduction to Complex Numbers - $i$ When people think of topics or facts that are "complex", they usually refer to a concept that requires multiple iterations of breaking it down into a simpler problem, then reassembling it into the frame of the original concept. However, complex **numbers** transcend this logic - instead being a figment of our imagination, requiring the use of abstract imagery and the utilisation of new systems to accommodate it. Still following? Let's jump in. First off, let's introduce our favourite complex number, $i$. If you haven't seen this before, don't fret - **it's simply the square root of minus one**. $i = \sqrt{ -1 }$ Note how this doesn't actually exist as a numerical value - hence the "imaginary" moniker we assign. Some complex numbers have both real and imaginary parts! This is known as the **complex cartesian form**, where numbers are written as shown: $z = 1 + 2i$ >[!Success]- Cartesian Additional Intuition >We can take a distinct cartesian plane for complex numbers where the y axis is given as a function of complex numbers and the x axis denotes a real number as shown: > >![[Pasted image 20230801150313.png]] > >As a result, a complex number anywhere on the plane can be expressed as a linear displacement value in the horizontal real plane and the vertical imaginary plane. ## Arguments and Moduli Another common way to write don't like them :( $z = rcis\theta$ $z^{*} = r (\cos \theta - i \sin \theta) $ # Complex Conjugates We can denote the *inverse* of a complex number as a *conjugate* of the original number, as shown: >[!Example]- Conjugate Example > >Given that there is a complex number $z$ with the value $1 + 2i$ find the complex conjugate of the number. > >**Solution:** > >Note how the conjugate is the *inverse* of the original number. This means that the **difference of two squares** of the number should yield a real value. Finding the conjugate becomes changing the sign of the negative to yield: >$z^* = 1-2i$ > >Simple as that! It is imperative that this knowledge sticks! Very, very important moving on, especially if you want to find the ranges of real/non-real values for imaginary numbers. # Quadratics with Complex Numbers Notice how GCSE questions told you to find solutions with *real* roots. Unfortunately for us, this means that they've neglected the cases where the discriminant is below zero for the sake of simplicity, leaving us with an understanding of This changes today! For a quadratic expression with a discriminant $b^2 - 4ac$ less than 0, we can use the quadratic theorem to find the complex roots of the polynomial - as seen in the example we've provided! So, take a quadratic expression $x^2 + 1$. This isn't a difference of two squares by any metric - meaning that our roots are complex. Using the quadratic formula, we get: $-\frac{b\pm\sqrt{ b^2-4ac }}{2a} = \frac{\sqrt{ -4 }}{2}$ The square root of $-4$ is 2i. Therefore, we can simplify the expression to give us both our roots: $\frac{\sqrt{ -4 }}{2} = \pm i$ We've found our roots! This allows us to find an equation for the expression: $x^2+1 = (x+i)(x-i)$ Note that **this actually means that complex roots must be conjugates of each other.** We can actually extend this to cubic, quartic, quintic [^1] equations and beyond! For this example, we'll stick to a cubic equation. >[!Success]- Chapter 4-3 Question 17 >Let p(x) be $x^3 + 4x^2 + 9x + 36$. >a) Show that p(3i) = 0. >b) Hence solve the equations p(x) = 0. > >**Solution a)** > >Factor theorem question! Remember that substituting the value in should give us 0 for the value to be a factor. >$(3i)^3 + 4(3i)^2 + 9(3i) + 36 = -27i - 36 + 27i + 36 = 0$ >This proves it! > >**Solution b)** >The above question shows us that one of the roots of the equation is $(x-3i)$. Given that complex roots form conjugates of each other (as given by the q) another root of the equation will be $(x+3i)$. To find the last root, we must therefore divide the equation by the quadratic formed by these roots, namely: >$(x-3i)(x+3i) = x^2 + 9$ >$\frac{x^3 + 4x^2 + 9x + 36}{x^2 + 9} = x+9$ [^1]: Note that this would be very, very difficult to find! As the saying goes, there is no quintic theory, so this is going to be a lot of trial and error. You have been warned! # De Moivre's Theorem De Moivre's Theorem is given by the identity: $(r\cos \theta + ri\sin \theta)^n = r^n\cos n\theta + r^ni\sin n\theta = r^ne^{i\theta n} = r^n cis \, n\theta$ Yeah i forgot how he got this lemme relearn and get back to ya >[!Success]- De Moivre's Theorem Proof by Induction >:( >[!Abstract]- De Moivre's Theorem Empirical Formula # Basic Trig Identities - Complex Numbers Now that we've got everything 'basic' down, it's time to create a launchpad for the next step - that being through *trigonometric identities*! De Moivre's theorem states that $e^{i \theta} = \cos \theta + i \sin \theta$. Note that this means that $e^{-i \theta} = \cos \theta - i \sin \theta$. This gives us the possibility of creating identities from this - can you see it? >[!Success]- The cos$\theta$ formula. >This one shouldn't be to difficult - not meaning to condescend! Add the two identities above to get: >$e^{i \theta} + e^{-i \theta} = 2 \cos \theta$ >So therefore: >$\cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2}$ Great! Remember to click on the spoiler - <span style="color:rgb(166, 227, 161)">green means important.</span> We can do the opposite to get an identity for $\sin \theta$, subtracting instead of adding: $\sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i}$ # Example Problems # What's Next? You're done with basic complex numbers! The next step would be to familiarise yourself with [[Polar Coordinates (Maths)|Polar Coordinates]] so that the next part will be a breeze! To get back to the main maths page, go here: [[Maths Contents Page]]