OSA: Complex Analysis

Question 1

Which of the following functions is holomorphic on \(\mathbb{C}\setminus \{0\}\)?

function	holomorphic?	explanation
\(\frac{\bar z}{\vert z \vert^2}\)	yesno	Since \(\|z\|^2 = z\cdot\bar z\) this is just \(\frac{1}{z}\) if \(z \not = 0\).
\(\frac{\bar z}{1+ \vert z \vert ^2}\)	yesno	The function \(\frac{\bar z}{1 + \|z\|^2}\) reverses orientation (for \(z\) close to 1, this function is approximately \(\frac{\bar z}{2}\)), so it can not be holomorphic.
\(\frac{\bar z}{z}\)	yesno	Since the identity function \(z\) is holomorphic, if \(\frac{\bar z}{z}\) was holomorphic then so would \(\bar z = z \cdot \frac{\bar z}{z}\) be.
\(\frac{1+z}{z}\)	yesno	Being the quotient of two holomorphic functions, this function is holomorphic.

Question 2

Compute the complex path integral \(\int_\gamma f(z)\,\mathrm{d}z\) for the following choices of \(f\) and \(\gamma\). In each case, the path \(\gamma\) is parameterised as a function \([0,1] \to \mathbb{C}\).

question	answer	result	explanation
\[\begin{split}f(z) &= \frac{i\bar z}{\pi},\\ \gamma(t) &= e^{2\pi it}\end{split}\]			By definition, \[\begin{split} \int_\gamma f(z)\,\mathrm{d}z &= \int_0^1 f\big(\gamma(t)\big)\cdot \gamma'(t)\,\mathrm{d}t \\ &= \frac{i}{\pi}\int_0^1 e^{-2\pi it} \cdot 2\pi i e^{2\pi it}\,\mathrm{d}t\\ &= -2 \end{split}\]
\[\begin{split} f(z) &= z^3,\\ \gamma(t) &= te^{it}\cos\big(\frac{\pi}{2}t\big) + (1-t)e^{t^2}\end{split} \]			Since \(z^3\) is holomorphic on all of \(\mathbb{C}\), by Cauchy's Theorem we may replace \(\gamma\) by another curve with the same endpoints \(\gamma(0) = 1\) and \(\gamma(1) = 0\). Choosing \(\eta(t) = 1-t\) we obtain \[\begin{split} \int_\gamma z^3\,\mathrm{d}z &= \int_\eta z^3\,\mathrm{d}z \\ &= \int_0^1 \eta(t)^3\cdot \eta'(t)\,\mathrm{d}t \\ &= \int_0^1 -(1-t)^3\,\mathrm{d}t\\ &= \left[\frac{(1-t)^4}{4}\right]_0^1 = -\frac{1}{4} \end{split}\]
\[\begin{split} f(z) &= \frac{i\cos(z^2)}{\pi z^5},\\ \gamma(t) &= e^{2\pi it} \end{split}\]			The function \(\frac{\cos(z^2)}{z^2}\) is holomorphic on \(\mathbb{C}\setminus\{0\}\) and has a pole of degree 2 at 0. Since \(\gamma\) winds exactly once around the pole, by the residue theorem we get \[\frac{1}{2\pi i}\int_\gamma f(z)\,\mathrm{d}z = c_{-1}, \] where \(c_{-1}\) is the coefficient of \(z^{-1}\) in the Laurent-expansion of \(f\). But \[ \frac{\cos(z^2)}{z^5} = \left(z^{-5} - \frac{1}{2}z^{-1} + \frac{1}{24}z^3 + \cdots\right), \] so \(c_{-1} = 2\pi i\cdot \frac{i}{\pi}\cdot(-\frac{1}{2}) = 1\).

Question 3

We consider the function \(f:\mathbb{C} \to \mathbb{C}\) with \[ f(a+bi) = \sin a + bi. \]

Determine the set of points \(z \in \mathbb{C}\) at which \(f\) is complex differentiable. Where is it holomorphic?
Calculate \(\int_\gamma f(z)\,\mathrm{d}z\) with \(\gamma(t) = t-it^2\), \(t \in [0,\pi]\).

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