Complex Numbers

What a complex number is

A complex number is a number of the form $z=x+iy$, where $x$ and $y$ are real numbers and $i$ is the imaginary unit, defined by the single property $i^2=-1$. The number $x$ is the real part of $z$, written $\mathrm{R}\mathrm{e}z$, and $y$ is the imaginary part, written $\mathrm{I}\mathrm{m}z$. Both of them are ordinary real numbers: for $z=3+4i$ the real part is 3 and the imaginary part is 4, not $4i$.

Every real number is a complex number with $y=0$, and a number with $x=0$ such as $5i$ is called purely imaginary. Real and imaginary numbers live inside one system, and that is the point of the topic: $x^2+1=0$ has no real solution, but it has the two complex solutions $z=i$ and $z=-i$.

Two complex numbers are equal exactly when both their real parts are equal and their imaginary parts are equal. So $a+bi=c+di$ means $a=c$ and $b=d$, and $3+iy=x-2i$ forces $x=3$ and $y=-2$. That one fact does all the work in 'compare real and imaginary parts' questions, and it is the engine of the square-root method later on this page.

Adding, subtracting, multiplying and dividing in Cartesian form

Addition and subtraction are done part by part: add the real parts, add the imaginary parts. $(3+4i)+(2-7i)=5-3i$ and $(3+4i)-(2-7i)=1+11i$. Nothing surprising happens and no $i^2$ appears.

Multiplication is expansion followed by $i^2=-1$. $(3+4i)(2-7i)=6-21i+8i-28i^2=6-13i+28=34-13i$. Every term containing $i^2$ becomes real, which is why a product of two complex numbers has no $i^2$ left in it once you have finished.

Division is multiplication by the conjugate of the denominator, top and bottom. To work out $\frac{5+3i}{2-i}$, multiply above and below by $2+i$: the denominator becomes $(2-i)(2+i)=4+1=5$, a real number, and the numerator becomes $(5+3i)(2+i)=10+5i+6i+3i^2=7+11i$. So the answer is $\frac{7+11i}{5}=\frac{7}{5}+\frac{11}{5}i$.

The trick works because $z{z}^{*}=x^2+y^2$, which is real. Multiplying a complex number by its own conjugate is the standard way to make a denominator real, and it is the same idea as rationalising a surd denominator such as $\frac{1}{2+\sqrt{3}}$.

• addition: $(3+4i)+(2-7i)=5-3i$
• multiplication: $(3+4i)(2-7i)=34-13i$, using $i^2=-1$
• division: multiply numerator and denominator by the conjugate of the denominator
• the denominator becomes $x^2+y^2$, a real number, so the answer is a genuine $x+iy$

The conjugate and conjugate pairs of roots

The conjugate of $z=x+iy$ is $z^{*}=x-iy$: the same real part, the imaginary part negated. The syllabus notation is $z^{*}$, and you will also meet $\bar{z}$ in textbooks. The conjugate of $3+4i$ is $3-4i$, the conjugate of a real number is itself, and $(z^{*}{)}^{*}=z$.

Conjugates turn division into real arithmetic because $z{z}^{*}=(x+iy)(x-iy)=x^2+y^2=|z{|}^2$. That identity is worth learning on its own: it gives you the modulus without drawing a triangle, and it is the reason a complex denominator becomes real.

The theorem the syllabus wants: if a polynomial equation has real coefficients, then any non-real root occurs together with its conjugate. If $1+3i$ is a root then $1-3i$ is also a root, so a cubic with real coefficients and one non-real root must have a third, real, root.

The theorem also tells you what to factorise. The conjugate roots $1+3i$ and $1-3i$ have sum 2 and product $1^2+3^2=10$, so they are the roots of the quadratic $z^2-2z+10=0$. Dividing the cubic by that quadratic leaves a linear factor whose root you can read off, and the sum of all the roots does the same job without any long division.

The Argand diagram

A complex number $z=x+iy$ is drawn as the point $(x,y)$: the horizontal axis is the real axis, the vertical axis is the imaginary axis, and the picture is called an Argand diagram. The number $3+4i$ is the point $(3,4)$, the number $-2-i$ is the point $(-2,-1)$, and the origin represents $0$.

Lengths and angles in the diagram are the modulus and the argument, which is why the diagram is the right picture for this topic. The point $z^{*}$ is the mirror image of $z$ in the real axis, and $-z$ is the point obtained by rotating $z$ through $\pi$ about the origin.

The geometrical effects of the operations are worth knowing in words, because a question may ask you to describe them: adding $w$ translates the point $z$ by the vector that represents $w$; multiplying by $i$ rotates a point through $\frac{\pi }{2}$ anticlockwise about the origin; multiplying by a general $w$ rotates through $\mathrm{\backslash arg}w$ and enlarges by $|w|$; and dividing by $w$ rotates through $-\mathrm{\backslash arg}w$ and shrinks by $|w|$.

Modulus and argument

The modulus of $z=x+iy$ is $|z|=\sqrt{x^2+y^2}$, the distance from the origin to the point $(x,y)$ - Pythagoras on the Argand diagram. The argument is the angle from the positive real axis to the line joining the origin to $z$, measured anticlockwise and written $\mathrm{\backslash arg}z$.

The argument is not unique: adding or subtracting $2\pi$ gives another angle describing the same point. Cambridge's usual interval is $-\pi <\theta \le \pi$, and the interval $0\le \theta <2\pi$ is sometimes more convenient. Answers may use either unless the question says which, but a question that asks for the argument 'in the interval $-\pi <\theta \le \pi$' means exactly that.

To find an argument, first find the reference angle $\alpha$ with the real axis from $\tan \alpha =\left|\frac{y}{x}\right|$, then use the quadrant of the point $(x,y)$ to convert it. A calculator's ${\tan }^{-1}$ gives an angle in the first or fourth quadrant only, so it cannot be trusted on its own.

For $z=-1+i\sqrt{3}$: $|z|=\sqrt{1+3}=2$, the point is in the second quadrant, and the reference angle is ${\tan }^{-1}\left(\frac{\sqrt{3}}{1}\right)=\frac{\pi }{3}$, so $\mathrm{\backslash arg}z=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$.

Polar form, and multiplying and dividing there

Writing $z$ as $x+iy$ is Cartesian form. Writing it as $r(\cos \theta +i\sin \theta )$ is polar form, where $r=|z|$ and $\theta =\mathrm{\backslash arg}z$. The shorter notation $r{e}^{i\theta }$ means the same thing and is used constantly in the papers, so $1+i=\sqrt{2}(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})=\sqrt{2}{e}^{i\pi /4}$.

Polar form is what makes multiplication and division easy. Expanding the product of $r_1(\cos {\theta }_1+i\sin {\theta }_1)$ and $r_2(\cos {\theta }_2+i\sin {\theta }_2)$ and using the addition formulae for $\cos ({\theta }_1+{\theta }_2)$ and $\sin ({\theta }_1+{\theta }_2)$ shows that the moduli multiply and the arguments add: $z_1{z}_2=r_1{r}_2(\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2))$ so $|z_1{z}_2|=|z_1||z_2|$ and $\mathrm{\backslash arg}(z_1{z}_2)=\mathrm{\backslash arg}{z}_1+\mathrm{\backslash arg}{z}_2$.

For division the arguments subtract: $\frac{z_1}{z_2}=\frac{r_1}{r_2}(\cos ({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2))$ so $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$ and $\mathrm{\backslash arg}\left(\frac{z_1}{z_2}\right)=\mathrm{\backslash arg}{z}_1-\mathrm{\backslash arg}{z}_2$.

For example, $z_1=2(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})$ and $z_2=3(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})$ give $|z_1{z}_2|=6$ and $\mathrm{\backslash arg}(z_1{z}_2)=\frac{\pi }{3}+\frac{\pi }{6}=\frac{\pi }{2}$, so $z_1{z}_2=6(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2})=6i$. Convert back to Cartesian form only at the end, when the question asks for $x+iy$.

Finding the square roots of a complex number

The two square roots of $5+12i$ are the two complex numbers whose square is $5+12i$. The method is to suppose the root is $a+bi$ with $a$ and $b$ real, square it, and compare real and imaginary parts with the target.

$(a+bi{)}^2=a^2-b^2+2abi$, so comparing with $5+12i$ gives $a^2-b^2=5$ and $2ab=12$. That is two equations in two unknowns, one of them quadratic, and solving it by substitution produces a cubic in $a^2$ - slow and easy to get wrong.

The shortcut is the modulus. $|(a+bi{)}^2|=|a+bi{|}^2$, so $|a+bi{|}^2=|5+12i|=13$, which gives the equation $a^2+b^2=13$. Adding and subtracting this with $a^2-b^2=5$ gives $a^2=9$ and $b^2=4$ immediately, so $a=\pm 3$ and $b=\pm 2$.

The remaining equation $2ab=12$ is positive, so $a$ and $b$ have the same sign, which leaves the two roots $3+2i$ and $-3-2i$, that is $\pm (3+2i)$. Both must be given: a square-root question asks for two numbers, and one of them alone is an incomplete answer.

Check: $(3+2i{)}^2=9+12i+4i^2=5+12i$. The negative root squares to the same thing because $(-w{)}^2=w^2$.

Loci in the Argand diagram

A locus question describes all the points $z$ that satisfy a condition, and each standard condition has a shape. The trick is to read the algebra as a statement about a distance or an angle before doing anything else.

$|z-a|=k$ says the distance from $z$ to the fixed point $a$ is $k$: a circle with centre $a$ and radius $k$. $|z-a|<k$ is the inside of that circle, and $|z-a|\le k$ is the inside together with the circle itself.

$|z-a|=|z-b|$ says $z$ is the same distance from $a$ as from $b$: the perpendicular bisector of the line segment joining $a$ and $b$. That is a straight line, and its Cartesian equation comes from equating the two squared distances.

$\mathrm{\backslash arg}(z-a)=\alpha$ says the line from $a$ to $z$ makes an angle $\alpha$ with the real axis: a half-line starting at $a$ and going outwards at that angle, with $a$ itself not included, because the argument of 0 is undefined.