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Messages - Alexander Elzenaar

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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.

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