Question

Can someone check if this is ok.
I need to find the locus of points x^2+y^2=ax in a complex plane after the 1/z transformation.

Answer 1

My answer is x=1/a

Answer 2

what is meant by 1/z transformation

Answer 3

This are the how:
|z|^2=aRe(z) ------->|1/z|^2=aRe(1/z)-------->1/|z|^2 =aRe(conj(z)/|z|^2)--------->1/x^2+y^2=ax/x^2+y^2-----> x=1/a

Answer 4

inverse transformation

Answer 5

can you confirm @eigenschmeigen?

Answer 6

i havent done anything like this before, is the method to substitute Z = 1/z?

Answer 7

it's not really a method, it just that every point that befor was z now become 1/z. So the curve of z will transform in somethnig. The question is , what?

Answer 8

can i do this then:

X + iY = 1/(x+iy)

X = x/(x^2 + y^2)

Y = -y/(x^2+y^2)

so x = X(x^2+y^2)

y = -Y(x^2+y^2)

now substitute in to get a new curve?

Answer 9

yes, that's what i did

Answer 10

ok sorry i couldn't tell really with all those arrows xD

so:

X^2 + Y^2 = aX/(x^2+y^2)

is that right?

Answer 11

hmm...
X^2 + Y^2 = aX/(x^2+y^2)^2 ?? Tht's what i get

Answer 12

by your method

Answer 13

nah i think:

X^2(x^2+y^2)^2 + Y^2(x^2+y^2)^2 = aX(x^2+y^2)


then

X^2 + Y^2 = aX/(x^2+y^2)

Answer 14

X = x/(x^2 + y^2)

Y = -y/(x^2+y^2)
so:
X^2 = x^2/(x^2 + y^2)^2

Y^2 = -y^2/(x^2+y^2)^2

so:
X^2+Y^2=x^2+y^2/(x^2 + y^2)^2=ax/(x^2 + y^2)^2

Answer 15

you're probably right, this transformation method always confuses me, im going to go away and learn it properly

Answer 16

thx for trying anyways,....)

Answer 17

@TuringTest help and give some good explain plz

Answer 18

I'm sorry, I know very little of complex analysis :(

Answer 19

k, thx

Answer 20

\[x^2+y^2=ax\]\[\left(x-\frac{a}{2}\right)^2+y^2=\left(\frac{a}{2}\right)^2\]\[z=\frac{a}{2}\left(1+e^{i\phi}\right)=\frac{a}{2}\left(1+\cos\phi+i\sin\phi\right)=\frac{a}{2}\left(2\cos^2{\frac{\phi}{2}}+i2\cos{\frac{\phi}{2}}\sin{\frac{\phi}{2}}\right)=\]\[=a\cos{\frac{\phi}{2}}\left(\cos{\frac{\phi}{2}}+i\sin{\frac{\phi}{2}}\right)=a\cos{\frac{\phi}{2}}e^{i\phi/2}\]\[\frac{1}{z}=\frac{1}{a}\frac{e^{-i\phi/2}}{\cos{\phi/2}}=\frac{1}{a}\left(1-i\tan{\frac{\phi}{2}}\right)\]

Answer 21

nice... Thnx a lot

Answer 22

would that be a vertical line?

Answer 23

yes, that is vertical line x=1/a