Complex functions - loci in the complex plane

A transformation, T, maps the point P from the z plane to Q in the w plane, defined below.

$\displaystyle w = \frac{2-iz}{z}, z \in \mathbb{C}, z\neq 0$

What is the locus of the point P, if P is mapped to Q which has locus $arg(w)=\frac{ \pi }{3}$.

I got $\displaystyle arg \left(\frac{z+2i}{z} \right )=\frac{5 \pi}{6}$.
Can someone just quickly check this and tell me what answer they got. Would be appreciated.

well it is correct but is there a reason why you want it like this?

What do you mean? The form that it is in?

yes

I think it probably the easiest to sketch in this form.

it is

I think it probably the easiest to sketch in this form.

Hi, sorry to hijack your thread but i have a question for you both

So i was trying to sketch $\displaystyle \displaystyle arg \left(\frac{z+2i}{z} \right )=\frac{5 \pi}{6}$ but give up and went to wolfham alpha to see it, only to find it was a line

usually my method is to pick some values for the argument and find where the lines meet. An example below;

(note: i will quote angles in degrees rather than radians, just because i cba latexing it, i know its wrong and im sorry :spank

I would do arg(z) = 0 therefore arg(z-2i) = -45, draw the lines and "X" where they cross
arg(z) = 60 therefore arg(z-2i) = 15 and do the same etc until I see the part circle which i can effectively dot to dot



however unless i am being really dumb this method cant be applied to OPs question, so what other (and probably better) method is there for these questions?

thanks, and once again sorry for hijacking OP

these are arcs of circles

yep.. is this it?


but is there a better method, my current method seems very unreliable

Hi, sorry to hijack your thread but i have a question for you both

So i was trying to sketch $\displaystyle \displaystyle arg \left(\frac{z+2i}{z} \right )=\frac{5 \pi}{6}$ but give up and went to wolfham alpha to see it, only to find it was a line

usually my method is to pick some values for the argument and find where the lines meet. An example below;

(note: i will quote angles in degrees rather than radians, just because i cba latexing it, i know its wrong and im sorry :spank

I would do arg(z) = 0 therefore arg(z-2i) = -45, draw the lines and "X" where they cross
arg(z) = 60 therefore arg(z-2i) = 15 and do the same etc until I see the part circle which i can effectively dot to dot



however unless i am being really dumb this method cant be applied to OPs question, so what other (and probably better) method is there for these questions?

thanks, and once again sorry for hijacking OP

yep.. is this it?


but is there a better method, my current method seems very unreliable

Look up the "arc of circles" video On examsolutions under complex loci

You will start at -2i, then draw an arc of a circle (minor arc as angle is greater than $\pi/2$) anticlockwise to the origin, o+oi.

the geometric approach is pants, but in exams will be adequate
I was educated to use algebra to obtain the equation of the circle, and then get the arc from algebraic constraints

...

Sounds like you got it, yep.

wtf that's so much easier than my method, thanks a lot

You understand why we start at (0,0) in your case though, right? If it was arg(z -(a+ib)) - arg(blah) you'd start from (a,b).

if this is right then im all good