Complex Variables Questions and Answers

Answer: C

\(|f'(-1 + i)|^{2} = 8\).

For an analytic function \(f'(z) = u_{x} + i\,v_{x}\). From the given parts:

\[ \begin{aligned} u_{x} &= y^{3} - 3x^{2}y \\ v_{x} &= x^{3} - 3xy^{2} \end{aligned} \]

(The Cauchy-Riemann equation \(u_{x} = v_{y}\) checks: \(v_{y} = y^{3} - 3x^{2}y\).) At \(x = -1\), \(y = 1\):

\[ \begin{aligned} u_{x} &= 1 - 3 = -2,\quad v_{x} = -1 + 3 = 2 \\ |f'|^{2} &= (-2)^{2} + 2^{2} = 8 \end{aligned} \]

The values 1, 7 and 10 come from sign slips in these two derivatives.

Answer: A

The function is \(f(z) = z|z|\).

At \(z = 0\) the difference quotient is \(\frac{f(h) - f(0)}{h} = |h| \to 0\), so \(f'(0) = 0\) exists. Elsewhere write \(r = \sqrt{x^{2} + y^{2}}\), \(u = xr\), \(v = yr\). Then

\[ \begin{aligned} u_{x} &= r + \tfrac{x^{2}}{r}, \quad v_{y} = r + \tfrac{y^{2}}{r} \\ u_{y} &= \tfrac{xy}{r}, \quad v_{x} = \tfrac{xy}{r} \end{aligned} \]

The Cauchy-Riemann equations \(u_{x} = v_{y}\) and \(u_{y} = -v_{x}\) need \(x^{2} = y^{2}\) and \(xy = 0\), which together force \(x = y = 0\). So \(z|z|\) is differentiable only at the origin.

\(\sin z\) and \(e^{-z^{2}}\) are entire, differentiable everywhere. \(e^{1/z}\) is differentiable at every \(z \ne 0\) but has an essential singularity at 0, so it is not even continuous there: the reverse of what is asked.

Answer: A

The correct statement is \(f\) is periodic.

\(e^{z + 2\pi i} = e^{z}e^{2\pi i} = e^{z}\), so \(e^{z}\) is periodic with the purely imaginary period \(2\pi i\), even though the real exponential \(e^{x}\) is not periodic.

Writing \(e^{z} = e^{x}(\cos y + i\sin y)\) settles the others. The modulus is \(|f| = e^{x}\), which equals 1 only on the imaginary axis. The argument is \(y + 2n\pi\), not \(y \pm n\pi\): adding an odd multiple of \(\pi\) would flip the sign of f. And since f is periodic, 'f is not periodic' is false.

Answer: B

The value is \(f(1) = 1 + 5i\).

Look for an analytic \(f'(z)\) whose real part is \(3x^{2}-4y-3y^{2}\). Since \(\text{Re}(3z^{2}) = 3x^{2}-3y^{2}\) and \(\text{Re}(4iz) = -4y\):

\[ f'(z) = 3z^{2} + 4iz + ic \]

for a real constant c, and f'(0) = 0 gives c = 0. Integrating:

\[ \begin{aligned} f(z) &= z^{3} + 2iz^{2} + K \\ f(i) &= -i - 2i + K = 0 \Rightarrow K = 3i \\ f(1) &= 1 + 2i + 3i = 1 + 5i \end{aligned} \]

Once f'(z) is fixed, f(i) = 0 fixes K, so f(1) has only one possible value. Dropping K would give 1 + 2i, which is not offered; 4 + 2i, 4 - 2i and 1 - i do not come from any f with this derivative that vanishes at i.

Answer: C

Both P and Q are true, and Q implies P.

Write \(u = x^{3}-4xy^{2}\) and \(v = 4x^{2}y-2y^{3}\). Then:

\[ \begin{aligned} u_{x} &= 3x^{2}-4y^{2}, & v_{y} &= 4x^{2}-6y^{2} \\ u_{y} &= -8xy, & v_{x} &= 8xy \end{aligned} \]

The second Cauchy-Riemann equation \(u_{y} = -v_{x}\) always holds, but \(u_{x} = v_{y}\) needs \(x^{2} = 2y^{2}\). On the x-axis (y = 0) that means x = 0, so the equations fail at every other point of the x-axis: Q is true.

Analyticity at a point needs the equations to hold throughout a neighbourhood of it. Every neighbourhood of (0, 0) contains points of the x-axis where they fail, so f is not analytic at the origin: P is true, and this argument derives P from Q, so Q implies P.

Answer: A

Complex differentiability at an isolated point is too weak to be useful; analyticity demands it on a whole open disc about the point. That extra requirement is what buys the powerful results -- infinite differentiability, a convergent Taylor series, and Cauchy's theorem -- none of which follows from differentiability at one point alone

Answer: D

Requiring the derivative to be the same whether the limit is taken along the real or the imaginary direction gives

\[ \frac{\partial u}{\partial x} \;=\; \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} \;=\; -\frac{\partial v}{\partial x}. \]

The minus sign belongs to the second pair only; putting it on the first, or omitting it altogether, describes no analytic function

Answer: C

Here \( u = x \) and \( v = -y \), so \( u_{x} = 1 \) while \( v_{y} = -1 \):

\[ u_{x} \;=\; 1 \;\neq\; -1 \;=\; v_{y}. \]

The first Cauchy-Riemann equation fails at every point, so the conjugate is nowhere analytic despite being continuous everywhere. Continuity is no guarantee of complex differentiability

Answer: B

The equations are necessary, and they become sufficient once the four partial derivatives are also continuous at the point. Without that extra hypothesis there are functions satisfying them at the origin yet not differentiable there, so the equations alone cannot be treated as a complete test

Answer: A

Differentiating the Cauchy-Riemann equations and adding eliminates \( v \):

\[ u_{xx} + u_{yy} \;=\; v_{yx} - v_{xy} \;=\; 0. \]

So both the real and the imaginary parts are harmonic, and this is what links complex analysis to two-dimensional potential problems in fluid flow and electrostatics