How does $f(z) = ic^{(1-z)}$ transform a certain region D on the complex plane

Question

The region is $ D = \{ z \in \mathbb{C}: Re(z) > 1; 0 < Im(z) < \frac{\pi}{2} \}$ and the transform is $f(z) = ic^{(1-z)}$ where $c$ is a constant.

I'm not sure where to begin as I've never come across a transform where $z$ is the power (asides from $exp(z)$) and I while I know that $z+1$ and $z-1$ move the plane a distace $1$ towards the right and the left, I'm not sure how $1-z$ behaves.

Thanks in advance.

Answer 1

Sorry I don't know how to format Is my first post on stack.exchange Hope it helps

f(z) = ic^(1-z)

• ie^((1-z)ln(c))

• ie^(ln(c) - (x+iy)ln(c))

• ie^(ln(c) - xln(c) -iyln(c))

• ie^(ln(c^(1-x)) -iyln(c))

• ic^(1-x).(cos(yln(c)) - i.sin(yln(c))

• u = c^(1-x).sin(yln(c))

• v = c^(1-x).cos(yln(c))

Other approach can be to take log on both sides

As x approaches infinity f(z) approaches origin

When u take y = 0 And let x move from 1 to infinity v goes from 1 to 0 for c>=1