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Quiz-3 / Re: Q3 TUT 0301

« 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 $.

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MAT334--Lectures & Home Assignments / Re: Section 1.4 Example 8

« on: September 20, 2018, 10:27:24 PM »

We simply need to show that the absolute value of $(1/n)\bigl(\cos(n\pi/4) + i\sin(n\pi/4)\bigr) $ tends to zero as $ n \to \infty $. Note first that $ \lvert (1/n)\bigl(\cos(n\pi/4) + i\sin(n\pi/4)\bigr)\rvert = 1/n $ (the trig functions vanish due to the Pythagorean identity).

I claim that for every $\varepsilon > 0 $ there exists some $ N \in \mathbb{N} $ such that for all $ n > N $ we have $ \lvert 1/n \rvert < \varepsilon $. Let such a $ \varepsilon $ be given; then by the Archimedian property of the real numbers, there exists a natural number $ N $ such that $ 1/\varepsilon < N $. If we pick any $ n > N $, then $ 1/\varepsilon < n $, and $ 1/n < \varepsilon $; we are done. (Note: I am justified in dropping the absolute value bars because everything is positive.)

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MAT334--Misc / Broken link on Quercus

« on: September 10, 2018, 08:55:20 PM »

The direct link from the Quercus page to the syllabus doesn't work ("The requested URL /courses/mat334h1/20189/outlines.html was not found on this server."). The link on the actual course page does work, though.