Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz3 => Topic started by: Victor Ivrii on October 12, 2018, 06:18:16 PM

Show that $w = \cos(z)$ maps the strip $\{z=x+yi\colon 0 < x < \pi\}$ both onetoone and onto the region obtained by deleting from the plane the two rays $(\infty, 1]$ and $[1, \infty)$. Draw both domains.
Hint: Use equalities $\cos (z)=\cos(\piz)$, and $\overline{\cos(z)} = \cos(\bar{z})$.

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 $.

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]

Alexander , you need to prove that you will get the region described
Yifei, what exactly you are trying to prove?