11. If \(f(z)=u(x, y)+i x y\) is analytic then
Explanation for Question 11: For \(f(z) = u(x, y) + iv(x, y)\) with \(v(x, y) = xy\), using the CR equation, we find \(v_{x} = y\) and \(v_{y} = x\). This leads us to conclude that \(u = \frac{x^{2}}{2} - \frac{y^{2}}{2}\). Thus, the true option is only \(c\).
12. Let \(f(z)=\frac{z}{|z|}\) when \(z \neq 0\) and \(f(z)=0\) when \(z=0\). Then
Explanation for Question 12: Considering \(f(z) = \begin{cases} \frac{z}{|z|} & \text{if } z \neq 0 \\ 0 & \text{if } z = 0 \end{cases}\), for \(z_{n} = \frac{1}{n} + 0i \rightarrow 0\), we find that \(f(z_{n}) \rightarrow 1 \neq f(0)\). Hence, \(f\) is discontinuous at \(z = 0\) as it does not have a limit at \(0\).
13. At \(z=0\), the function \(f(z)=z^{2} \bar{z}\)
Explanation for Question 13: By checking the complex form of the CR equation, i.e., \(\frac{\partial f}{\partial \bar{z}} = 0\), we find that \(\frac{\partial }{\partial \bar{z}} z^{2} \bar{z} = z^{2} = 0\) only at \(z = 0\). Therefore, \(f\) satisfies the CR equation at \(0\) and is also a polynomial in \(z,\bar{z}\). This implies that \(f\) is differentiable at \(z = 0\), but the CR equation is not satisfied in any neighborhood of \(z = 0\), hence \(f\) is not analytic.
14. The function \(f: \mathbb{C} \rightarrow \mathbb{C}\) defined by \(f(z)=|z|\) is
Explanation for Question 14: Considering \(f(2) = |z|\), we observe that \(f\) is a real-valued function. A real-valued function is analytic if and only if it is constant. However, \(f\) is not constant, implying that \(f\) cannot be analytic anywhere.
15. \(\alpha \neq 1\) is a complex number such that \(\alpha^{5}=1\). Which of the following is true?
Explanation for Question 15: Given \(\alpha^{5} = 1\) and \(\alpha \neq 1\), we find that \(\alpha^{5} - 1 = 0\), leading to \((\alpha-1)(1+\alpha+\alpha^{2}+\alpha^{3}+\alpha^{4}) = 0\). However, it is also true that \(\alpha^{5} = 1\) implies \(\alpha^{4} = \frac{1}{\alpha}\). Further analysis reveals that \(\alpha^{5} = 1\) implies \(re^{i5\theta} = 1\) for \(\alpha = re^{i\theta}\), leading to the values of \(\theta = \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}\), or \(\frac{8\pi}{5}\). Hence, options \(b\), \(c\), and \(d\) are not true.
16. Suppose \(f=u+i v\) is defined in some neighborhood of \(z_{0}\) and that \(u, v, u_{x}, u_{y}, v_{x}, v_{y}\) are continuous and satisfy the Cauchy-Riemann equations at \(z_{0}\). Then read the following statements.
1. \(f\) is analytic at \(z_{0}\)
2. \(z_{0}\) is a regular point of \(f\)
3. \(f^{\prime}\left(z_{0}\right)\) exists
Explanation for Question 16: All statements are true, as it is a necessary and sufficient condition.
17. Which of the following functions is bounded as \(z\) varies over the complex plane?
Explanation for Question 17: Using the fact that a non-constant entire function cannot be bounded, since \(a\), \(b\), and \(c\) are all non-constant and entire, they are unbounded.
18. If \(z_{1}, z_{2}\) are two complex numbers such that \(\left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right|\) then it is necessary that
Explanation for Question 18: Analyzing \(z_{1} \cdot z_{2} = 0\), we find that \(z_{1} = 0\) or \(z_{2} = 0\) (verify). Also, \(\left|z_{1}+z_{2}\right| = \left|z_{1}\right| + \left|z_{2}\right|\). Furthermore, if \(z_{1} = \lambda z_{2}\), then \(\left|z_{1}+z_{2}\right| = (\lambda + 1)\left|z_{2}\right| = \left|\lambda z_{2}\right| + \left|z_{2}\right| = \left|z_{1}\right| + \left|z_{2}\right|\). Hence, the strongest condition is that \(\left|z_{1}+z_{2}\right| = \left|z_{1}\right| + \left|z_{2}\right|\).
19. The values of \(i\) where \(i\) is the square root of -1 is
Explanation for Question 19: Using the polar form of \(i\), we find \(i = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} = e^{i\frac{\pi}{2}}\). By solving the equations, we find \(v = x^{2} - y^{2} + 2y - c\).
20. The function \(f(z)=|z|^{2}+i \bar{z}+1\) is differentiable at
Explanation for Question 20: By applying the CR Equation in the complex form, \(\frac{\partial f}{\partial \bar{z}} = \frac{\partial}{\partial \bar{z}}(z \bar{z} + i \bar{z} + 1) = z + i = 0\). This implies that \(f\) can only be differentiable at \(z = -i\), since the CR equation is satisfied only at \(-i\). Using the same logic as in Question 13, \(f\) is differentiable at \(-i\).
0 Comments