Arithmetic of complex numbersEdexcel International A Level Further Maths: Revision notes

Treat $\mathrm{i}$ like an algebraic symbol and replace $\mathrm{i}^2$ by $-1$. To add or subtract, combine real parts and imaginary parts separately: $(3+2\mathrm{i})+(1-4\mathrm{i})=4-2\mathrm{i}$. To multiply, expand the brackets: $(3+2\mathrm{i})(1-4\mathrm{i})=3-12\mathrm{i}+2\mathrm{i}-8\mathrm{i}^2=11-10\mathrm{i}$. Powers follow by repeated multiplication: $(1+2\mathrm{i})^2=-3+4\mathrm{i}$ and $(1+2\mathrm{i})^3=(-3+4\mathrm{i})(1+2\mathrm{i})=-11-2\mathrm{i}$. A number is purely imaginary when its real part is $0$. For $z_1+\lambda z_2=(3+\lambda)+(2-4\lambda)\mathrm{i}$ to be purely imaginary, $3+\lambda=0$, so $\lambda=-3$.

To find a quotient, multiply the numerator and denominator by the conjugate of the denominator. This makes the denominator real, because $(c+d\mathrm{i})(c-d\mathrm{i})=c^2+d^2$. $\dfrac{3+2\mathrm{i}}{1-4\mathrm{i}}=\dfrac{(3+2\mathrm{i})(1+4\mathrm{i})}{(1-4\mathrm{i})(1+4\mathrm{i})}=\dfrac{3+12\mathrm{i}+2\mathrm{i}-8}{17}=\dfrac{-5+14\mathrm{i}}{17}$. So the real part is $-\frac{5}{17}$ and the imaginary part is $\frac{14}{17}$. Always give the final answer in the form $p+q\mathrm{i}$ by splitting the fraction.

Work in steps and keep the form $a+b\mathrm{i}$ at each stage. To find $\dfrac{z^2}{z^*}$ with $z=1+2\mathrm{i}$: first $z^2=-3+4\mathrm{i}$ and $z^*=1-2\mathrm{i}$. Then $\dfrac{-3+4\mathrm{i}}{1-2\mathrm{i}}\times\dfrac{1+2\mathrm{i}}{1+2\mathrm{i}}=\dfrac{-3-6\mathrm{i}+4\mathrm{i}-8}{5}=-\dfrac{11}{5}-\dfrac{2}{5}\mathrm{i}$. Reuse earlier results where the question says 'hence': here the result of $z^2$ saves expanding again.

A complex number $a+b\mathrm{i}$ is the position vector $\binom{a}{b}$. Adding complex numbers adds the vectors, so if $A$ and $B$ represent $w_1$ and $w_2$, the point $C$ for $w_1+w_2$ completes the parallelogram $OACB$. For $w_1=4+\mathrm{i}$ and $w_2=1+3\mathrm{i}$, $C$ represents $5+4\mathrm{i}$. The difference $w_2-w_1$ represents the vector from $A$ to $B$, so the distance $AB=|w_2-w_1|=|-3+2\mathrm{i}|=\sqrt{13}$.

Multiplying by a positive real number $k$ is an enlargement of scale factor $k$ about $O$. Multiplying by $\mathrm{i}$ is a rotation of $90^\circ$ anticlockwise about $O$: $\mathrm{i}(4+3\mathrm{i})=-3+4\mathrm{i}$ turns the point $(4,3)$ to $(-3,4)$, keeping the same distance $5$ from $O$. Dividing by $\mathrm{i}$ rotates $90^\circ$ clockwise, since $\frac1{\mathrm{i}}=-\mathrm{i}$. Because $z_2=\mathrm{i}z_1$ has the same modulus as $z_1$ and is perpendicular to it, the points $O$, $z_1$, $z_1+z_2$, $z_2$ form a square. Quotients such as $\frac{z_1}{z_1+z_2}=\frac12-\frac12\mathrm{i}$ have modulus $\frac{|z_1|}{|z_1+z_2|}=\frac{5}{5\sqrt2}=\frac{1}{\sqrt2}$, the ratio of the two distances from $O$.