Sketch the region onto which the sector 0<r<=2/3 and

0<= theta <= pie/4 is mapped by the transformations

1) f(z)=1/z

2)z^3

where z=re^(itheta)

I dont understand graphing in the complex plane.

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- Jan 28th 2010, 01:22 PMscubasteve123sketching complex
Sketch the region onto which the sector 0<r<=2/3 and

0<= theta <= pie/4 is mapped by the transformations

1) f(z)=1/z

2)z^3

where z=re^(itheta)

I dont understand graphing in the complex plane. - Jan 28th 2010, 02:44 PMshawsend
I'll do the first one. You have $\displaystyle f(z)=1/z=\frac{1}{r}e^{-it}$ right? Gotta' get that far first ok. So suppose I let $\displaystyle z=1 e^{it}$ and let t go from 0 to 2pi. Then $\displaystyle f(1 e^{it})$ is just a circle of radius one. Suppose I let $\displaystyle z=1/4 e^{it}$. Ok, then $\displaystyle f(1/4 e^{it})$ is then a circle of radius 4. Suppose I let $\displaystyle z=1/4 e^{it}$ but this time I only let t go from 0 to pi. Then that's half a circle but it's the lower half because of the negative exponent. How about if I let $\displaystyle z=1/4 e^{it}$ and let t go from 0 to $\displaystyle \pi/4$ then $\displaystyle f(1/4 e^{it})$ is 1/8 of a circle of radius 4 in the fourth quadrant going from 0 to -pi/4 and the smaller I let r, the larger the radius of that circle under $\displaystyle f(z)$ because it's 1/r. Ok, what about all of these 1/8 circles with r going from 0 to 2/3? Wouldn't that be the pi/4 sector in the fourth quadrant with inner radius 3/2 but extending to infinity? Like the plot below which shows the region 0<r<2/3 and 0<=t<=pi/4 on the left, and the image of that region under f(z)=1/z on the right except the red part extends out to infinity and I just chopped it.