FP2.3 - Further complex numbers

Euler's relation represents a rotation through a real angle $\theta$ as a complex exponential: $e^{i\theta}=\cos\theta+i\sin\theta$. Its real and imaginary parts are the coordinates of the corresponding point on the unit circle.

The second relation is obtained by replacing $\theta$ with $-\theta$ and using that cosine is even while sine is odd. Adding the two equations eliminates the imaginary terms; subtracting them eliminates the cosine terms.

For example, $$e^{3i\theta}+e^{-3i\theta}=2\cos3\theta,$$ while $$e^{3i\theta}-e^{-3i\theta}=2i\sin3\theta.$$ These conversions let a trigonometric expression be handled with laws of indices and then converted back.

De Moivre's theorem says that multiplying a complex number in polar form $n$ times multiplies its argument by $n$ and raises its modulus to the power $n$. For every integer $n$,$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta).$$More generally, $[r(\cos\theta+i\sin\theta)]^n=r^n[\cos(n\theta)+i\sin(n\theta)]$.

For positive integers, the result follows by induction. The case $n=1$ is immediate. If it holds for $n=k$, multiplication by $\cos\theta+i\sin\theta$ and the angle-addition formulae give $\cos((k+1)\theta)+i\sin((k+1)\theta)$. For $n=0$, both sides equal $1$. For negative $n$, take the reciprocal of the positive-power result; the reciprocal of $\cos\phi+i\sin\phi$ is $\cos\phi-i\sin\phi$. This proves the theorem for every integer $n$.

For instance, equating imaginary parts when $n=3$ gives $$\sin3\theta=3\sin\theta\cos^2\theta-\sin^3\theta=3\sin\theta-4\sin^3\theta.$$ In the reverse direction, Euler's forms give $$\cos^2\theta=\frac{1+\cos2\theta}{2}.$$ Thus the same theorem connects multiple angles with powers in both directions.

The $m$ values have equal modulus $R^{1/m}$ and arguments separated by $2\pi/m$, so they form a regular $m$-gon centred at the origin. Substitute one root into the original equation to check the modulus and multiplied argument.

On an Argand diagram, $|z-a|$ is the distance from the point $z$ to the fixed point $a$, while $\arg(z-a)$ is the directed angle of the displacement from $a$ to $z$. Translate each complex condition into distance or angle geometry before sketching.

For an exact equation, set $z=x+iy$ and write each modulus as a squared distance. For example, $$|z|=2|z-3|$$ becomes $$x^2+y^2=4[(x-3)^2+y^2],$$ so $$(x-4)^2+y^2=4.$$ The locus is therefore the circle with centre $(4,0)$ and radius $2$.

For an inequality, first draw its equality boundary, test a point not on that boundary to choose the correct side, and use a solid boundary for $\le$ or $\ge$. For an argument condition, also respect the stated argument range so that the correct ray or arc is selected.

A transformation assigns each permitted point $z$ in the source plane a point $w$ in the image plane. To find the image of a locus, connect $z=x+iy$ and $w=u+iv$ through the transformation, substitute into the source condition, and simplify until the result is a condition on $u$ and $v$.

For example, let $$w=\frac{z-1}{z+1},\qquad z\ne-1.$$ If $z$ lies on the imaginary axis, then $z=iy$ for real $y$, so $$|w|=\frac{|iy-1|}{|iy+1|}=1.$$ The image lies on the unit circle. Solving $z=(1+w)/(1-w)$ shows that $w=1$ has no finite preimage, so the image is the unit circle with that point excluded.

For a non-degenerate fractional linear transformation, $ad-bc\ne0$. If $c\ne0$, the source value $z=-d/c$ is a pole and is excluded; the inverse also reveals any missing image value. Lines and circles commonly map to lines or circles, but the algebra and exclusions decide which one in a particular problem.

After obtaining the image equation, check it with one easy source point: transform that point directly and verify that its $w$-coordinates satisfy the new equation. This catches sign errors introduced while rationalising or completing the square.