Author Topic: Q3 TUT 0301  (Read 911 times)

Victor Ivrii

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Q3 TUT 0301
« on: October 12, 2018, 06:18:16 PM »
Show that $w = \cos(z)$ maps the strip $\{z=x+yi\colon 0 < x < \pi\}$ both one-to-one and onto the region obtained by deleting from the plane the two rays $(\infty, -1]$ and $[1, \infty)$. Draw both domains.

Hint: Use equalities $\cos (-z)=-\cos(\pi-z)$, and $\overline{\cos(z)} = \cos(\bar{z})$.

Alexander Elzenaar

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Re: Q3 TUT 0301
« Reply #1 on: October 14, 2018, 04:58:19 PM »
Here is a proof for injectiveness:-

Let us call the strip $ S $. Consider $ w, z \in S $; we want to show that $ \cos w - \cos z = 0 $ implies $ w = z $. Since the normal sum and product identities for the real functions carry over to the complex functions, we have $ 0 = -2\sin \frac{w + z}{2} \sin \frac{w - z}{2} $. Thus either $ 0 = \sin \frac{w + z}{2} $ or $ 0 = \sin \frac{w - z}{2} $. The zeroes of the complex sine function occur only on the real line, so either $ \frac{w + z}{2} = n\pi $ or $ \frac{w - z}{2} = n\pi $ for some $ n \in \mathbb{Z} $.

Suppose then that $ w + z = 2n\pi $. Since they add to give a real number, $ w $ and $ z $ are complex conjugates of each other. Thus their sum is just the sum of their real parts. Their real parts lie in the interval $ (0,\pi) $ by assumption; thus the real part of $ w + z $ lies in the interval $ (0, 2\pi) $ and hence there is no $ n $ satisfying the condition (i.e. there is no number of the form $ 2n\pi $ in the interval $ (0,2\pi) $. So this case is not possible.

It follows then that the only possibility is $ w - z = 2n\pi $. If $ w = x + iy $ and $ z = u + iv $, the only possible way for $ w - z $ to be real is for $ y $ to equal $ v $; so they have the same imaginary part. On the other hand, we have that the real part of $ w - z $ lies in the interval $ (-\pi, \pi) $ (since $ 0 < x < \pi $ and $ -\pi < u < 0 $); the only number of the form $ 2n\pi $ in this interval is zero, and so $ w $ and $ z $ have the same real part. Combining these two observations, $ w = z $.

Yifei Wang

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Re: Q3 TUT 0301
« Reply #2 on: October 15, 2018, 10:39:25 PM »
I have a different approach to this question (without using the hint).


Let W as cos(x)cosh(y)-isin(x)sinh(y)
When X = 0:
sin(x) = 0, cos(x) = 1
W = cosh(y)

When X = pi
sin(x) = 0, cos(x) = -1
W =- cosh(y)

As we know cosh(y) in [1, limits) , and sin(x) in [1, -1].
Therefore, we can conclude y in (-limits, limits) and x in [-1, 1]

Victor Ivrii

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Re: Q3 TUT 0301
« Reply #3 on: October 16, 2018, 06:57:14 AM »
Alexander , you need to prove that you will get the region described

Yifei, what exactly you are trying to prove?