### Author Topic: What's the answer of FE q2 d.e.f?  (Read 1772 times)

#### xinliu

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##### What's the answer of FE q2 d.e.f?
« on: December 11, 2018, 05:50:14 PM »
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

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##### Re: What's the answer of FE q2 d.e.f?
« Reply #1 on: December 11, 2018, 08:12:08 PM »
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

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##### Re: What's the answer of FE q2 d.e.f?
« Reply #2 on: December 11, 2018, 08:17:28 PM »
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

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##### Re: What's the answer of FE q2 d.e.f?
« Reply #3 on: December 11, 2018, 08:28:39 PM »
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

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##### Re: What's the answer of FE q2 d.e.f?
« Reply #4 on: December 11, 2018, 08:31:32 PM »
Can you explain how you figured it out?

#### JUNJING FAN

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##### Re: What's the answer of FE q2 d.e.f?
« Reply #5 on: December 11, 2018, 08:39:41 PM »
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.