I have an impression that the OP wanted to get some answers about the result without having to compute too many points. Below is an attempt, at least for a):

If I remember correctly rational complex functions always map straight lines and circles to straight lines and circles. But [imath]\lim_{z\rightarrow i} f(z) = \infty[/imath], which means that the image of [imath]f[/imath] must be not a circle but a straight line. Now one can compute a couple of points and figure out the exact line.

I am sure that you mean [imath]\lim_{z\rightarrow -i}f(z)[/imath]

And when you analyze this limit you will get

[imath]\lim_{z\rightarrow -i^{+}}f(z) = +\infty[/imath]

And

[imath]\lim_{z\rightarrow -i^{-}}f(z) = -\infty[/imath]

which is an indication that you have a line.

But that idea will not be so obvious if you are not familiar with that specific mapping.

It is like when you are in a calculus test and you are given this integral:

[imath]\displaystyle \int_{-\pi}^{\pi} x^3\ln\left(x^2\ln x^4\right)\sin^2x \ dx[/imath]

How many students will fail to answer this integral and say we did not study this?!

And

How many students will think outside the box and will realize that the integrand is

**odd**?

which is an indication that:

[imath]\displaystyle \int_{-\pi}^{\pi} x^3\ln\left(x^2\ln x^4\right)\sin^2x \ dx = 0[/imath]

But you have to know that stuff. (

**odd** thing.)