Complex numbers

Complex numbers and their parts Paper 3

The equation $x^2=-1$ has no real solution. Mathematicians define a new number $\mathrm{i}$ with ${\mathrm{i}}^2=-1$, and then every quadratic equation has solutions: $\sqrt{-16}=\sqrt{16}\sqrt{-1}=4\mathrm{i}$.

A complex number is $z=x+\mathrm{i}y$ with $x$ and $y$ real. $x$ is the real part, $\mathrm{Re}z$, and $y$ (not $\mathrm{i}y$) is the imaginary part, $\mathrm{Im}z$. So for $z=3-4\mathrm{i}$, $\mathrm{Re}z=3$ and $\mathrm{Im}z=-4$. The conjugate of $z$ is $z^{\times }=x-\mathrm{i}y$: the sign of the imaginary part changes.

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. One complex equation therefore gives two real equations, which is how you find unknown real numbers in a complex equation: expand, collect real and imaginary parts, and equate each.

Arithmetic in Cartesian form Paper 3

Add and subtract by combining real parts and imaginary parts separately. Multiply by expanding the brackets, then replace ${\mathrm{i}}^2$ by $-1$:
$$(2+3\mathrm{i})(4-\mathrm{i})=8-2\mathrm{i}+12\mathrm{i}-3{\mathrm{i}}^2=11+10\mathrm{i}$$
A number times its conjugate is real: $z{z}^{\times }=(x+\mathrm{i}y)(x-\mathrm{i}y)=x^2+y^2$. That is the key to division: multiply the top and bottom by the conjugate of the bottom, exactly as you rationalise a surd denominator.
$$\frac{5+\mathrm{i}}{2-3\mathrm{i}}=\frac{(5+\mathrm{i})(2+3\mathrm{i})}{(2-3\mathrm{i})(2+3\mathrm{i})}=\frac{7+17\mathrm{i}}{13}=\frac{7}{13}+\frac{17}{13}\mathrm{i}$$
Show every line of a multiplication or division: the marks are for the working.

Roots of polynomial equations Paper 3

A quadratic with a negative discriminant has two complex roots: $z^2-4z+13=0$ gives $z=\frac{4\pm \sqrt{-36}}{2}=2\pm 3\mathrm{i}$. They are conjugates of each other, and that is no accident.

For a polynomial equation with real coefficients, non-real roots occur in conjugate pairs. The reason: conjugating a sum or product conjugates each piece, and real coefficients are their own conjugates, so $(\mathrm{p}(w){)}^{\times }=\mathrm{p}(w^{\times })$. If $\mathrm{p}(w)=0$, then $\mathrm{p}(w^{\times })=0^{\times }=0$.

So if $w$ is a root, $w^{\times }$ is too, and $(z-w)(z-w^{\times })=z^2-2\mathrm{Re}(w)z+|w{|}^2$ is a real quadratic factor. For a cubic or quartic with one complex root given, write down its conjugate, form this quadratic factor, and divide (or compare coefficients) to find the rest.

The Argand diagram, modulus and argument Paper 3

An Argand diagram shows $z=x+\mathrm{i}y$ as the point $(x,y)$, with the real part across and the imaginary part up.

• The modulus $\left|z\right|=\sqrt{x^2+y^2}$ is the distance from $O$ to the point.
• The argument $\arg z$ is the angle from the positive real axis to the line from $O$ to the point, anticlockwise positive. It is usually taken in $-\pi <\theta ⩽\pi$ (the principal argument); $0⩽\theta <2\pi$ is sometimes more convenient, and either is accepted unless a question says otherwise.

Find the argument from a sketch, not from the calculator alone. ${\tan }^{-1}\frac{y}{x}$ only gives angles between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$, so in the second and third quadrants it must be adjusted. For $z=-3+4\mathrm{i}$ (second quadrant), the angle with the negative real axis is ${\tan }^{-1}\frac{4}{3}=0.927$, so $\arg z=\pi -0.927=2.21$.

$|-3+4\mathrm{i}|=5$ and $\arg (-3+4\mathrm{i})=2.21$: measured from the positive real axis, not from the nearest axis.

Polar and exponential forms Paper 3

With $r=\left|z\right|$ and $\theta =\arg z$, the real and imaginary parts are $r\cos \theta$ and $r\sin \theta$, so
$$z=r(\cos \theta +\mathrm{i}\sin \theta )\equiv r{\mathrm{e}}^{\mathrm{i}\theta }$$
The exponential form is a compact way of writing the polar form. These forms make multiplication and division easy:
$$z_1{z}_2=r_1{r}_2{\mathrm{e}}^{\mathrm{i}({\theta }_1+{\theta }_2)},\frac{z_1}{z_2}=\frac{r_1}{r_2}{\mathrm{e}}^{\mathrm{i}({\theta }_1-{\theta }_2)}$$
So $\left|z_1{z}_2\right|=\left|z_1\right|\left|z_2\right|$ and $\arg (z_1{z}_2)=\arg z_1+\arg z_2$, with the matching results for division. Why: expand $(\cos {\theta }_1+\mathrm{i}\sin {\theta }_1)(\cos {\theta }_2+\mathrm{i}\sin {\theta }_2)$; the real part is $\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2=\cos ({\theta }_1+{\theta }_2)$ and the imaginary part is $\sin ({\theta }_1+{\theta }_2)$, by the addition formulae.

After adding or subtracting arguments, add or subtract $2\pi$ if needed to bring the answer into the interval asked for.

Square roots of a complex number Paper 3

To find the square roots of $z$, let a root be $a+\mathrm{i}b$ with $a$ and $b$ real, and set $(a+\mathrm{i}b{)}^2=a^2-b^2+2ab\mathrm{i}$ equal to $z$. Equating real and imaginary parts gives two equations. Substitute $b$ from the second into the first to get a quadratic in $a^2$; only the positive value of $a^2$ is allowed, because $a$ is real. There are always two roots, and each is the negative of the other. Show every step: the working carries the marks.

The geometry of the operations Paper 3

Each operation has a picture on the Argand diagram:

• conjugating reflects the point in the real axis;
• adding $w$ translates by the vector representing $w$, so $z+w$ is found by the parallelogram law, as with vectors;
• subtracting: $z-w$ is the vector from $w$ to $z$, so $|z-w|$ is the distance between the points;
• multiplying by $w=r{\mathrm{e}}^{\mathrm{i}\alpha }$ enlarges by scale factor $r$ and rotates anticlockwise through $\alpha$ about $O$; in particular, multiplying by $\mathrm{i}$ is a rotation through $\frac{\pi }{2}$;
• dividing by $w$ scales by $\frac{1}{r}$ and rotates through $-\alpha$.

Loci in the Argand diagram Paper 3

Reading $|z-a|$ as "the distance from $z$ to $a$" turns equations into shapes:

• $|z-a|=k$ is the circle with centre $a$ and radius $k$; $|z-a|<k$ is the inside, drawn with a dashed boundary;
• $|z-a|=|z-b|$ is the perpendicular bisector of the line joining $a$ and $b$;
• $\arg (z-a)=\alpha$ is the half-line from $a$ (not including $a$) at angle $\alpha$ to the positive real direction.

Write $z=x+\mathrm{i}y$ to get a Cartesian equation: $|z-(3+4\mathrm{i})|=2$ becomes $(x-3{)}^2+(y-4{)}^2=4$.

Greatest and least values come from the geometry. For points on the circle $|z-c|=k$, $\left|z\right|$ runs from $\left|c\right|-k$ to $\left|c\right|+k$ (along the line through $O$ and the centre), and the greatest argument is where a line from $O$ touches the circle: $\arg c+{\sin }^{-1}\frac{k}{\left|c\right|}$.

The circle $|z-2-2\mathrm{i}|=1$ and the half-line $\arg (z-1)=\frac{\pi }{4}$.