STEP II 1997 question 5 solution

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If , then if we equate real and imaginary parts of the equation we get:

Since , then and thus . Whence , and so . That is,

If is constant, say , then

Thus

So the corresponding locus in the x-y plane is simply a circle centered at the origin of radius .

On the other hand, if is constant, say , then

Thus

That is, the corresponding locus in the x-y plane is a straight line through the origin of gradient , i.e. it makes an angle of with the positive x-axis.

So, for the last bit, the following subsets do the trick:

Solution by dvs