complex numbers
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.
$z + \bar z$ means you add up a complex number and its conjugate and the result must be higher than $0$.
Thus $(a + bi) + (a - bi) = 2a$. Sketch the graph $2*\hspace{2 pt}Re(z)$.
If you simply interpret what it means to say that $\arg(z-1)<\arg(z-i)$ geometrically, then it shouldn't be too hard to visualize the region. In each of the regions below, the argument of the point ($z$) minus the number (either $1$ or $i$) simply the angle from the dashed line to the arrow. It's easy to see that this angle is smaller from the complex number $1$ in exactly the blue regions.
Note that I'm assuming a branch cut at $\pi$ in the argument function so that, in the horizontal strip, $\arg(z-1)$ is definitely negative while $\arg(z-i)$ is positive.
Best Answer
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 45ยบ angle, or $\alpha=\pi/4$
And the radius is $r=\sqrt{2}(3+\sqrt{3})$