### Author Topic: Q3 TUT 0301  (Read 905 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]