How does one go about solving $arg(\frac{z-2i}{z-6}) = \frac{1}{2}\pi$

This should give $$\frac{z-2i}{z-6} = bi$$

but solving that gives me $$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$

and substituting $$z$$ for $$x + yi$$ gives me $$x = \frac{-2b +6b^2}{1+b^2}$$ and $$y=\frac{-6b +2}{1+b^2}$$

And I have no clue how to continue now. The answer is suppose to be $$x = -by$$, but I have no clue how to get there (have I made a calculation error??)

EDIT: Thanks everyone!!

• You did no mistakes. just divide the last two equations and you will get your result Commented Sep 15, 2020 at 10:42

The expression means that z subtends an angle $$\frac{\pi}{2}$$ at the points $$2i$$ and $$6$$

Ponder upon the following visual

The green arrows show some more positions which are possible for $$z$$. Clearly, as it subtends $$90^o$$ at those points, so the locus of $$z$$ is a semicircle, the end point of hose diameter is $$(0,2)$$ and $$(6,0)$$

• Thanks, this really helps Commented Sep 15, 2020 at 11:13

Most of the time, polar form is better when dealing with complex numbers. Draw a circle with $$2i$$ and $$6$$ as the ends of its diameter. This diameter divide the circle into 2 parts, the lower half is the loci.

Hint: if $$AB$$ is the diameter of a circle and $$P$$ is on the circle, $$\angle APB =\pm \frac{\pi}{2}$$

As you said solving gets you here: $$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$

and finally, substituting gets u here: $$x = \frac{-2b +6b^2}{1+b^2}$$ and $$y=\frac{-6b +2}{1+b^2}$$

Just divide above two equations you get

$$x = \frac{-2b + 6b^2}{1 + b^2} = -\frac{2 - 6b}{1 + b^2}\cdot b = -by$$ $$\implies x+by=0$$

• oh wow, that's stupidly simple, I have no clue how I've missed that. Thanks!! Commented Sep 15, 2020 at 11:10

You have not made a mistake. Observe that $$x$$ is a multiple of $$y$$:

$$x = \frac{-2b + 6b^2}{1 + b^2} = -b \cdot \frac{2 - 6b}{1 + b^2} = -by$$

which is what the answer states.