My textbook(Year 12 Cambridge Extension 2) asked me this question:
For question (b), is there a more simple way of answering this question than this:
It really isn't that big but, I would expect something less cumbersome for question 4 in the exercise and I feel I may be not seeing an extremely simple solution. Thanks to anyone who can help :)
On
Rotating anticlockwise by $90^\circ$ in the complex plane can by done by multiplying by $i$. $B=A+C$. The angle $\widehat{BOC}=45^\circ$ so $B'$ is in the direction of $C$ but $\sqrt{2}$ times longer.
$C=(2+i)i=2i+i^2=-1+2i$
$B=A+C=(2+i)+(-1+2i)=1+3i$
$B'=C\sqrt{2}=(-1+2i)\sqrt{2}=-\sqrt{2}+2\sqrt{2}i$
I should think you're expected to do both parts with complex numbers rather than plane geometry, though. This is only slightly more complicated. We need to know that rotating anticlockwise by $45^\circ$ can be done by multiplying by $(1+i)/\sqrt{2}$.
$B'=(1+3i)(1+i)/\sqrt{2}=(1+i+3i+3i^2)/\sqrt{2}=(-2+4i)/\sqrt{2}=-\sqrt{2}+2\sqrt{2}i$
If you have learned Euler’s formula $e^{i\theta}=\cos\theta + i\sin\theta$, then there is a more general theorem:
multiply a complex number $z$ with $e^{i\theta}$, you will get the point rotated $\theta$ in radians counterclockwise.
And when you replace the $\theta$ here with $45^\circ$, or $\frac\pi 4$, you will get the number $\frac 1{\sqrt{2}}(1+i)$.
The proof of this theorem is given by the sum of angle identity of $\sin$ and $\cos$. If originally $z$ has a magnitude of $r$ and angle $\phi$ with the $x$-axis, then:
$$ \begin{align*} z_2 &= z_1e^{i\theta} \\ &= (r\cos(\phi)+r\sin(\phi)i)(\cos(\theta)+\sin(\theta)i) \\ &= r\cdot[(\cos(\phi)\cos(\theta)-\sin(\phi)\sin(\theta)) - (\cos(\phi)\sin(\theta)+\sin(\phi)\cos(\theta))i] \\ &= r\cos(\theta+\phi) + r\sin(\theta+\phi)i, \end{align*} $$
which represents the complex number with magnitude $r$ and angle $(\theta+\phi)$ between the $x$-axis.