What's the answer of FE q2 d.e.f?

[xinliu]

There are only answers of part a and part b and it seems that we cannot reply to the question anymore.
Can somebody please post the answer of part d e f?
Thanks

[JUNJING FAN]

There are only answers of part a and part b and it seems that we cannot reply to the question anymore.
Can somebody please post the answer of part d e f?
Thanks

same, I am stuck at proving the function to be 1 to 1.

[JUNJING FAN]

There are only answers of part a and part b and it seems that we cannot reply to the question anymore.
Can somebody please post the answer of part d e f?
Thanks

also, for the graphing of the domains, textbook section 1.5 question 26 refers to figure 1.26 on page 53 as the domain of the mapping function. In the question it states that the mapping is one-to-one, but from what I can see it is not one-to-one.

[JUNJING FAN]

There are only answers of part a and part b and it seems that we cannot reply to the question anymore.
Can somebody please post the answer of part d e f?
Thanks

never mind, it is one-to-one. I figured it out.

[Jeffery Mcbride]

Can you explain how you figured it out?

[JUNJING FAN]

Can you explain how you figured it out?

so refer to textbook page 50, the book provides a method. For x just follow the book, for y follow it until the end to prove that y1+y2=0, however the range of y is not limited to be positive in this question. In this case, we will have to expand:

cos(x+iy) = cosx coshy - i sinx sinhy

and realize that sinx is never 0, meaning sinhy will always be there, and sinhy is injective for a positive or negative y of same magnitude.