Misc. Math Practice Questions

The poles of \(f(z)\) are at \(z = i\) and \(z = -i\). Both poles lie inside the circle \(|z| = 2\). By the Residue Theorem, the integral is \(2\pi i\) times the sum of the residues inside the contour. The residue at \(z = i\) is \(\text{Res}(f, i) = \lim_{z \to i} (z-i) \frac{1}{(z-i)(z+i)} = \lim_{z \to i} \frac{1}{z+i} = \frac{1}{2i}\). The residue at \(z = -i\) is \(\text{Res}(f, -i) = \lim_{z \to -i} (z+i) \frac{1}{(z-i)(z+i)} = \lim_{z \to -i} \frac{1}{z-i} = \frac{1}{-2i}\). The sum of residues is \(\frac{1}{2i} + \frac{1}{-2i} = 0\). Therefore, the integral is \(2\pi i \cdot 0 = 0\).

Answer: A) 0

We know the geometric series formula \(\frac{1}{1-z} = \sum_{n=0}^\infty z^n\) for \(|z| < 1\). Differentiating term by term with respect to \(z\) gives \(\frac{1}{(1-z)^2} = \frac{d}{dz} \left( \sum_{n=0}^\infty z^n \right) = \sum_{n=1}^\infty n z^{n-1}\). Let \(m = n-1\), so \(n = m+1\). When \(n=1\), \(m=0\). Then \(\frac{1}{(1-z)^2} = \sum_{m=0}^\infty (m+1) z^m = \sum_{n=0}^\infty (n+1) z^n\) for \(|z| < 1\).

Answer: (D) \(\sum_{n=0}^\infty (n+1) z^n\)

DIY

Answer: (C) \(1/6\)

By Cauchy's Integral Formula, \(\oint_C \frac{f(z)}{z - z_0} dz = 2\pi i \, f(z_0)\) for \(f\) analytic inside \(C\) and \(z_0\) inside \(C\). Here \(f(z) = e^z\) and \(z_0 = 1\), which lies inside \(|z|=2\). So the integral equals \(2\pi i \cdot e^1 = 2\pi i e\).

Answer: (B) \(2\pi i e\)

Write \(\frac{1}{z(z-1)} = \frac{1}{z} \cdot \frac{1}{z-1} = \frac{1}{z} \cdot \frac{-1}{1-z}\). For \(|z| < 1\), \(\frac{1}{1-z} = \sum_{n=0}^\infty z^n\). Thus \(f(z) = -\frac{1}{z} \sum_{n=0}^\infty z^n = -\frac{1}{z} - 1 - z - z^2 - \cdots = -\frac{1}{z} - \sum_{n=0}^\infty z^n\).

Answer: (D) \(-\frac{1}{z} - \sum_{n=0}^\infty z^n\)

By the generalized Liouville theorem, if \(f\) is entire and \(|f(z)| \leq M|z|^n\) for large \(|z|\), then \(f\) is a polynomial of degree at most \(n\). Here \(n = 2\), so \(f\) must be a polynomial of degree at most 2.

Answer: (C) \(f\) is a polynomial of degree at most 2

The Cauchy-Riemann equations are a necessary condition for analyticity: \(u_x = v_y\) and \(u_y = -v_x\). These ensure the complex derivative \(f'(z)\) exists.

Answer: (A) \(u_x = v_y\) and \(u_y = -v_x\)

Apply Rouché's theorem with \(f(z) = z^5\) and \(g(z) = 3z + 1\). On \(|z| = 2\): \(|f(z)| = 32\) and \(|g(z)| \leq 3(2) + 1 = 7 < 32\). Since \(|f(z)| > |g(z)|\) on \(|z|=2\), by Rouché's theorem \(p(z) = f(z) + g(z)\) has the same number of zeros inside \(|z|=2\) as \(f(z) = z^5\), which has 5 zeros (counting multiplicity).

Answer: (E) 5

Since \(\sin z = z - \frac{z^3}{6} + \frac{z^5}{120} - \cdots\), we have \(\frac{\sin z}{z^4} = \frac{1}{z^3} - \frac{1}{6z} + \frac{z}{120} - \cdots\). The lowest power of \(z\) in the denominator is \(z^3\), so \(z = 0\) is a pole of order 3.

Answer: (B) 3

The Möbius transformation \(w = \frac{z - i}{z + i}\) maps the upper half-plane to the unit disk. To verify: when \(z = 0\) (on the real axis), \(w = \frac{-i}{i} = -1\), which is on the unit circle. When \(z = i\) (interior of upper half-plane), \(w = 0\), which is in the unit disk. This is the standard conformal map from \(\mathbb{H}\) to \(\mathbb{D}\).

Answer: (C) \(w = \frac{z-i}{z+i}\)

By the Maximum Modulus Principle, if \(f\) is analytic and non-constant on a connected open set \(\Omega\), then \(|f(z)|\) cannot attain its maximum at any interior point of \(\Omega\). Therefore, the maximum of \(|f|\) on a closed bounded region is attained only on the boundary.

Answer: (D)

Consider \(f(z) = \frac{1}{1+z^2} = \frac{1}{(z-i)(z+i)}\). Integrating over a semicircular contour in the upper half-plane, the pole inside is at \(z = i\). The residue there is \(\lim_{z \to i}(z-i)\frac{1}{(z-i)(z+i)} = \frac{1}{2i}\). By the residue theorem, \(\int_{-\infty}^{\infty} \frac{dx}{1+x^2} = 2\pi i \cdot \frac{1}{2i} = \pi\).

Answer: (A) \(\pi\)

The Schwarz Lemma states: if \(f : \mathbb{D} \to \mathbb{D}\) is analytic and \(f(0) = 0\), then \(|f(z)| \leq |z|\) for all \(z \in \mathbb{D}\) and \(|f'(0)| \leq 1\). Equality \(|f(z_0)| = |z_0|\) for some \(z_0 \neq 0\), or \(|f'(0)| = 1\), implies \(f(z) = e^{i\theta} z\) for some real \(\theta\).

Answer: (E)

The Laurent expansion of \(e^{1/z} = \sum_{n=0}^\infty \frac{1}{n! z^n} = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \cdots\) has infinitely many negative-power terms. A singularity with infinitely many principal part terms is an essential singularity. By the Casorati-Weierstrass theorem, near an essential singularity the function comes arbitrarily close to every complex value.

Answer: (B) Essential singularity

The principal logarithm \(\text{Log}(z) = \ln|z| + i\,\text{Arg}(z)\) uses the principal argument \(\text{Arg}(z) \in (-\pi, \pi]\). It is analytic everywhere except on the branch cut, which is the non-positive real axis \(\{x \leq 0\}\), where \(\text{Arg}(z)\) is discontinuous.

Answer: (C)

On a simply connected domain, every harmonic function \(u\) (satisfying \(\nabla^2 u = 0\)) has a harmonic conjugate \(v\) such that \(u + iv\) is analytic. This is a consequence of the fact that the domain has no holes, so the conjugate differential form is exact.

Answer: (A)

Since \(z=1\) is a pole of order 2, the residue is \(\lim_{z \to 1} \frac{d}{dz}\left[(z-1)^2 f(z)\right] = \frac{d}{dz}\left[\frac{z}{z+2}\right]_{z=1}\). Computing: \(\frac{d}{dz}\frac{z}{z+2} = \frac{(z+2) - z}{(z+2)^2} = \frac{2}{(z+2)^2}\). At \(z = 1\): \(\frac{2}{(1+2)^2} = \frac{2}{9}\).

Answer: (D) \(\dfrac{2}{9}\)

By Liouville's Theorem, the only bounded entire functions are constants. Among the choices, \(f(z) = \sin(1/2)\) is a constant (a real number), hence trivially entire and bounded. All other options are non-constant entire functions, which by Liouville's theorem must be unbounded.

Answer: (C)

In an annular region \(r < |z-z_0| < R\), an analytic function is represented by its Laurent series \(\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n\), which includes both non-negative and negative powers of \(z\). A Taylor series (only non-negative powers) is valid only in a disk, not a punctured annulus with a hole.

Answer: (B)

For \(z = x \in \mathbb{R}\), compute \(\left|\frac{x - i}{x + i}\right| = \frac{|x - i|}{|x + i|} = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} = 1\). So every real \(z\) maps to a point with \(|w| = 1\). Since Möbius transformations map circles/lines to circles/lines, and the real axis (a generalized circle) maps onto \(|w| = 1\). This is consistent with \(f\) being the standard map from the upper half-plane to the unit disk.

Answer: (D) The unit circle \(|w| = 1\)

Use the Euclidean algorithm: \(252 = 1 \cdot 180 + 72\), \(180 = 2 \cdot 72 + 36\), \(72 = 2 \cdot 36 + 0\). The last non-zero remainder is 36. Thus, \(\gcd(252, 180) = 36\).

Answer: (B) 36

For a number \(n = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}\), the number of positive divisors is \((e_1 + 1)(e_2 + 1) \cdots (e_k + 1)\). Here, \(n = 2^3 \cdot 3^2 \cdot 5^1\), so the number of divisors is \((3+1)(2+1)(1+1) = 4 \cdot 3 \cdot 2 = 24\).

Answer: (C) 24

DIY

Answer: (A) 47