Complex Numbers: Expressing a point on argand diagram In an Argand diagram, the loci
$$
\arg(z-2i) =\pi/6 \quad  \land \quad |z-3|=|z-3i|
$$
intersect at the point $P$. Express the complex number represented by $P$ in the form  $re^{i \alpha}$, where $e$ is exponential,  giving the exact value of $\alpha$ and the value of $r$ correct to $3$ significant figures.
Im still new to the topic. I sketched the loci required but it was useless. I literally dont know how to figure it out. Im in grade 12. 
 A: let me label these points in the argand diagram. let $A = 2i, B = 3i, C = 3, P = z, D = 3/2 + 3i/2$ is the midpoint of $BC.$ 
since $arg(AP) = \pi/6,$  we can write $$z = 2i + 2r(\cos \pi/6 + i \sin \pi 6) = r\sqrt 3+i(r+2)$$  where $r$ is a positive number. we need to find $r$ so that $DP$ is orthogonal to $BC$ this is equivalent to the slope of $DC$ being one.
that is $$\frac{r + 2 - 3/2}{r\sqrt 3 - 3/2} = 1 \to r(\sqrt 3 - 1) = 2 \to r = \frac 2{\sqrt 3 - 1}, z = \frac{2\sqrt 3}{\sqrt 3 - 1} + i\frac{2\sqrt 3}{\sqrt 3 - 1} $$
A: Looking at the second condition:
$$|z-3|^2=(x-3)^2+y^2$$
$$|z-3i|^2=x^2+(y-3)^2$$
If they are equal, that means
$$(x-3)^2+y^2=x^2+(y-3)^2$$
$$x^2-6x+9+y^2=x^2+y^2-6y+9$$
$$-6x+=-6y$$
$$\left(*\right)\space \space \space \space x=y$$
Now looking at the first condition: $Arg(z-2i)=\frac{\pi}{6}$ we get
$$tg\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{3}$$
$$tg[Arg(z-2i)]=\frac{Im(z-2i)}{Re(z-2i)}=\frac{x-2}{x}$$
$$\frac{x-2}{x}=\frac{\sqrt{3}}{3}$$
$$x=y=3+\sqrt3$$
$$z=(3+\sqrt3)+(3+\sqrt3)i$$
Since both coordinates are the same, the point is at a 45º angle, or $\alpha=\pi/4$
And the radius is $r=\sqrt{2}(3+\sqrt{3})$
A: To find point $P$, you are given that $\arg(z-2i)=\frac{\pi}{6}$, and $|z-3|=|z-3i|$. 
Let us denote $P=a+bi$, where $a$ and $b$ are real numbers, which we wish to find.
From the angle information, we have
$$\arctan\left(\frac{\Im(P)-2}{\Re(P)}\right)=\arctan\left(\frac{b-2}{a}\right)=\frac{\pi}{6}$$
and using the magnitude information $|P-3|=|P-3i|$ results in
$$(a-3)^2+b^2=a^2+(b-3)^2\Rightarrow a=b$$
Using this, we can find $a$ from the angle information:-
$$a=\frac{2}{1-\tan\left(\frac{\pi}{6}\right)}=4.732$$
So, $P=4.73(1+i)=6.69e^{i\frac{\pi}{4}}$, the angle being the arctan of $1$, which is $\pi/4$ radians exactly, and $r$ is expressed to 3 significant figures.
