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Strange complex numbers question

I'm working through some worksheets I've been sent by Oxford University to attempt before I get there, and there's this question on complex numbers I just can't make work:

Show that the mapping w = z +c/z,where z = x + iy, w = u + iv and c is a real number, maps the circle |z| = 1 in the z plane into an ellipse in the w plane and find its equation.

I've tried multiplying through by z, and multiplying the final term by z*/z* to make a real bottom, before equating real and imaginary bits. I've also tried finding u and v in terms of x and y, but I can't arrive at an equation for an ellipse.

Thanks for any help
(edited 8 years ago)
Reply 1
Avatar for Zacken
Zacken
22
Original post by Luke Kostanjsek
x


Haven't worked through it, but this may be helpful:

You know that x2+y2=1x^2 + y^2 = 1 and that w=x+cxx2+y2+i(y−cyx2+y2)=x+cx+i(y−cy)\displaystyle w = x + \frac{cx}{x^2 + y^2} + i \left(y - \frac{cy}{x^2 + y^2}\right) = x+ cx +i (y-cy)
(edited 8 years ago)
Original post by Zacken
Haven't worked through it, but this may be helpful:

You know that x2+y2=1x^2 + y^2 = 1 and that w=x+cxx2+y2+i(y−cyx2+y2)=x+cy+i(y−cy)\displaystyle w = x + \frac{cx}{x^2 + y^2} + i \left(y - \frac{cy}{x^2 + y^2}\right) = x+ cy +i (y-cy)


Thanks for the reply. I've tried something similar to that, and I've got it in all kinds of different forms, but I still can't find a solution. The answer has to be in the form:

u^2/a^2 + v^2/b^2 = 1, doesn't it? I've not a clue how I'd arrive at anything like that :/
Reply 3
Avatar for Zacken
Zacken
22
Original post by Luke Kostanjsek
Thanks for the reply. I've tried something similar to that, and I've got it in all kinds of different forms, but I still can't find a solution. The answer has to be in the form:

u^2/a^2 + v^2/b^2 = 1, doesn't it? I've not a clue how I'd arrive at anything like that :/


Okay, I got it!

You have the form u+iv=x+cx+i(y−cy)\displaystyle u + iv = x + cx + i(y-cy) as above. Comparing the real and imaginary parts - we have u=x(1+c)  ⟹  x=u1+c\displaystyle u = x(1+c) \implies x = \frac{u}{1+c} - do the same and isolate yy in terms of vv.

Now you know that x2+y2=1x^2 + y^2 = 1, and you have xx and yy in terms of uu and vv - so substitute 'em in.
Original post by Zacken
Okay, I got it!

You have the form u+iv=x+cx+i(y−cy)\displaystyle u + iv = x + cx + i(y-cy) as above. Comparing the real and imaginary parts - we have u=x(1+c)  ⟹  x=u1+c\displaystyle u = x(1+c) \implies x = \frac{u}{1+c} - do the same and isolate yy in terms of vv.

Now you know that x2+y2=1x^2 + y^2 = 1, and you have xx and yy in terms of uu and vv - so substitute 'em in.


Aargh! I got x and y in u and v, but I didn't think to substitute them into that equation! :facepalm:

Thank you so much, you're a genius :P
Reply 5
Avatar for Zacken
Zacken
22
Original post by Luke Kostanjsek
Aargh! I got x and y in u and v, but I didn't think to substitute them into that equation! :facepalm:

Thank you so much, you're a genius :P


Just a stroke of luck on my part seeing that.

Have fun at Oxford! I'm infinitely envious. :h:

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