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QUESTION: Let t e C and consider the map:

f(t)=1/t+i

Explain what happens to all horizontal/vertical lines under the map f. Writing down the equations of the images of the lines and also explain them geometrically.

I tried writing f(t)=t-i/t^2+1, but after that I'm stuck.

Thanks

ANSWER: Dave, I'm a little unsure about your notation and the way you've interpreted the equation. First of all, I believe t is a complex number rather than a real-valued parameter; is that right? Second, should the equation be written

f(t) = (1/t) + i or f(t) = 1/(t+i) ??

It looks like you assumed the 2nd since you tried to rid the denominator of an imaginary component (always a good idea!), but you didn't do it correctly. If t is complex then you want to calculate (for w = f(t) and t = x + iy where x and y are real )

w = 1/(t+i) = 1/(x+iy+i) = 1/[x+i(y+1)] = [x-i(y+1)]/[x^2+(y+1)^2].

From this, letting w = u+iv means that the real part u = x/[x^2+(y+1)^2] and the imaginary part v = (y+1)/[(x^2+(y+1)^2].

Cross-plotting these and figuring out what the curve looks like for x = a = constant and y = b = constant will answer your question.

On the other hand, if it is assumed

w = [ 1/(x+iy) ] +i = { [1+i(x+iy)]/(x+iy) }{ (x-iy)/(x-iy) } = { x/(x^2+y^2) } + i{1-y/(x^2+y^2)} so that

u = x/(x^2+y^2) and v = 1-y(x^2+y^2)

which has a nice symmetry, but ...

Please let me know the answers to the questions above and I'll take another crack at it.

Randy

---------- FOLLOW-UP ----------

QUESTION: Hi Randy,

Yes t is the complex number and the equation is f(t) = 1/(t+i).

Thank you

OK, so we have f(t) = w = 1/(t+i). From the previously derived expression for w, let y+1 = p for convenience, then the real and imaginary parts of the transform w are

u = x(x^2+p^2)^1/2 and v = -p/(x^2+p^2).

It can be seen that v = -(p/x)u or p = -xv/u

To see how a horizontal line maps under the transformation, set x = a = constant, then, we get

au^2 -u + av^2 = 0.

By completing the square, this becomes

(u-1/(2a))^2 + v^2 = 1/(4a^2)

which is a circle centered at 1/(2a) and radius (1/2a).

You can play the same game by letting y = b = constant and get more circles. Good luck.

Randy

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