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FP2-Complex numbers

The transformation T from the z-plane to the w-plane is given by: w=2i/z. Show that T maps |z-3i|=3 to a line in the w-plane, and give the Cartesian equation of this line. Could anyone help pls?? :smile:

Reply 1

808

I didn't do that well in FP2 but since you have no replies I'll try:

(this was wrong)

Not too sure about that but there you go

Reply 2

goldsilvy

You have to make z the subject. Then sub the equivalent of z into the equation

|z-3i|=3

. Then solve it through squaring both sides. Also

w=u + iv

.

EDIT: Nope, wait, something doesn't work in here :p:

Reply 3

808

|z-3i|=3

(z-3i)\left(z^*+3i\right)=9

z=x+i y

w=u+i v

z=\frac{2v+2i u}{|w|}

z^*=\frac{2v-2i u}{|w|}

\left(\frac{2v+2i u}{u^2+v^2}-3i\right)\left(\frac{2v-2i u}{u^2+v^2}+3i\right)=9

\frac{9 u^2-12 u+9 v^2+4}{u^2+v^2}=9

9\frac{|w|}{|w|}-\frac{12u}{|w|}+\frac{4}{|w|}=9

\Rightarrow u = \frac{1}{3}

edit: fixed it to agree with later post. I made a mistake!

Reply 4

Pheylan

w=\frac{2i}{z}\Rightarrow z=\frac{2i}{w}

z-3i=\frac{2i-3iw}{w}\Rightarrow |z-3i|=\frac{|2i-3iw|}{|w|}

3|w|=|2i-3iw|

3|u+iv|=2i-3i(u+iv)|\Rightarrow 3|u+iv|=|2i-3iu+3v|

3|u+iv|=|3v+i(2-3u)|

9(u^2+v^2)=9v^2+(2-3u)^2

9u^2=4-12u+9u^2

u=\frac{1}{3}

Reply 5

Anti-Derivative

Ah thanks all for your help :smile:

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