Home | Assessment | Notes | Index | Worksheets | Blackboard
Calculate \[ \left| \dfrac{2^{n+1}}{n+1} \Bigg/ \dfrac{2^n}{n} \right| = \dfrac{2n}{n+1} \to 2 \] as $n \to \infty$ so the radius of convergence is $\frac{1}{2}$.
Calculate \[ \left| \dfrac{(n+1)!}{n!} \right| = n+1 \to \infty \] as $n \to \infty$ so the radius of convergence is $0$.
Calculate \[ \left| \dfrac{(n+1)^p}{n^p} \right| = \left( \dfrac{n+1}{n} \right)^p \to 1 \] as $n \to \infty$ so the radius of convergence is $1$.
Calculate \[ \left| \dfrac{(-1)^{n+1}}{((n+1)!)^2 2(n+1)} \Bigg/ \dfrac{(-1)^n}{(n!)^2 2n} \right| = \dfrac{n}{n+1} \cdot \dfrac{1}{(n+1)^2} \to 0 \] as $n \to \infty$ so $R = \infty$.
If $R \ne S$ the radius of convergence will be $\min \{R,S\}$. Indeed, if $|z| < \min \{R,S \}$ then both power series converge, so their sum does as well, but if $\min\{R,S\} < |z| < \max\{S,R\}$ the sum cannot converge as then both series would.
If $R = S$ then the sum can have any radius of convergence $T \ge R$. Indeed, suppose $b_n = - a_n$ and fix a power series \[ \sum_{n=0}^\infty c_n z^n \] with radius of convergence $T \ge R$. Then \[ \sum_{n=0}^\infty (a_n + c_n) z^n \qquad \sum_{n=0}^\infty (b_n + c_n) \] both have radius of convergence $R$ by the previous paragraph, but their sum has radius of convergence $T$.
We calculate $f(z+h) = z^2 + 2zh + h^2 + z + h$ so that \[ \dfrac{f(z+h) - f(z)}{h} = \dfrac{2zh + h^2 + h}{h} = 2z + h + 1 \] and \[ \lim_{h \to 0} \dfrac{f(z+h) - f(z)}{h} = \lim_{h \to 0} 2z + h + 1 = 2z + 1 \] for all $z \in \C$.
We calculate \begin{align*} \dfrac{f(z+h) - f(z)}{h} & = \dfrac{\dfrac{1}{z+h} - \dfrac{1}{z}}{h} = \dfrac{\dfrac{-h}{(z+h)z}}{h} = \dfrac{-1}{(z+h)z} \end{align*} and \[ \lim_{h \to 0} \dfrac{f(z+h) - f(z)}{h} = \lim_{h \to 0} \dfrac{-1}{(z+h)z} = -\dfrac{1}{z^2} \] for all $z \in \C \setminus \{0\}$.
From the definition \[ \lim_{h \to 0} \dfrac{f(h) - f(0)}{h} = \dfrac{|h|^2}{h} = \lim_{h \to 0} \dfrac{h \overline{h}}{h} = \lim_{h \to 0} \overline{h} = 0 \] so the function has a derivative of zero at the origin.
Fix $z \in \C$. The limit \begin{align*} \lim_{h \to 0} \dfrac{f(z+h) - f(z)}{h} & = \lim_{h \to 0} \dfrac{(z+h)\overline{(z+h)} - z \overline{z}}{h} \\ & = \lim_{h \to 0} \dfrac{z\overline{h} + \overline{z} h + |h|^2}{h} \end{align*} does not exist because if $h = t$ is real we get \[ \lim_{t \to 0} \dfrac{zt + \overline{z} t + t^2}{t} = z + \overline{z} \] whereas if $h = it$ is imaginary we get \[ \lim_{t \to 0} \dfrac{zit - \overline{z} it + t^2}{it} = \overline{z} - z \] so $f$ is not differentiable at any non-zero $z$. It is therefore not holomorphic on any domain.
From the definition \begin{align*} (\partial_1 h)(x,y) & = \lim_{t \to 0} \dfrac{h(x+t,y) - h(x,y)}{t} \\ & = \lim_{t \to 0} \dfrac{2(x+t)y - 2xy}{t} = \lim_{t \to 0} 2y = 2y \end{align*} and \begin{align*} (\partial_2 h)(x,y) & = \lim_{t \to 0} \dfrac{h(x,y+t) - h(x,y)}{t} \\ & = \lim_{t \to 0} \dfrac{2x(y+t) - 2xy}{t} = \lim_{t \to 0} 2x = 2x \end{align*} as expected from rules of differentiation.
From \[ \frac{1}{z} = \frac{x-iy}{x^2 + y^2} = \frac{x}{x^2 + y^2} + i \frac{(-y)}{x^2+y^2} \] we get $u(x,y)$ and $v(x,y)$. The partial derivatives are \begin{align*} (\partial_1 u)(x,y) & = \dfrac{(x^2 + y^2) - 2x(x)}{(x^2 + y^2)^2} = \dfrac{y^2 - x^2}{(x^2 + y^2)^2} \\ (\partial_2 u)(x,y) & = \dfrac{-2xy}{(x^2 + y^2)^2} \end{align*} and \begin{align*} (\partial_1 v)(x,y) & = \dfrac{2xy}{(x^2 + y^2)^2} \\ (\partial_2 u)(x,y) & = \dfrac{-(x^2 + y^2) + 2y(y)}{(x^2 + y^2)^2} = \dfrac{y^2 - x^2}{(x^2 + y^2)^2} \end{align*} so the Cauchy-Riemann equations are satisfied.
For this function $u(x,y) = \sqrt{x^2 + y^2}$ and $v(x,y) = 0$. Thus \[ (\partial_1 u)(x,y) = \dfrac{x}{\sqrt{x^2 + y^2}} \qquad (\partial_2 u)(x,y) = \dfrac{y}{\sqrt{x^2 + y^2}} \] and $(\partial_1 v)(x,y) = (\partial_2 v)(x,y) = 0$ for all $(x,y) \ne (0,0)$. The Cauchy- Riemann equations are therefore not satisfied at any non-zero complex number.
Since the partial derivatives of $u$ do not even exist at $(0,0)$ the function $f$ is certainly not differentiable there.
We calculate \begin{align*} (\partial_1 u)(x,y) & = 3x^2 - 3y^2 \\ (\partial_2 u)(x,y) & = -6xy \\ (\partial_1 v)(x,y) & = 6xy \\ (\partial_2 v)(x,y) & = 3x^2 - 3y^2 \end{align*} and see that the equations are satisfied for all $(x,y)$.
From \[ u(x,y) + i v(x,y) = x^3 + 3i x^2 y - 3xy^2 - y^3 \] we see that \[ f(z) = z^3 = (x+iy)^3 \] has the above real and imaginary parts.
We write \[ \dfrac{1}{(x+iy)^4} = \dfrac{(x-iy)^4}{(x^2+y^2)^4} = \dfrac{x^4 - 4ix^3 y - 6x^2y^2 + 4ixy^3 + y^4}{(x^2 + y^2)^4} \] so that $u$ and $v$ are the real and imaginary parts of $f(z) = \frac{1}{z^4}$ on $\C \setminus \{0\}$. Since $f$ is holomorphic on that domain, the Cauchy-Riemann equations are satisfied for all $z$ therein.
We calculate \begin{align*} \partial_1(\partial_1 u) + \partial_2(\partial_2 u) & = \partial_1(\partial_2 v) + \partial_2 (-\partial_1 v) \\ & = \partial_1 (\partial_2 v) - \partial_2 (\partial_1 v) = 0 \end{align*} by Clairaut's theorem.
From an earlier problem we have \[ u(x,y) = x^3 - 3x y^2 \qquad v(x,y) = 3x^2 y - y^3 \] and can calculate that \begin{align*} \partial_1 (\partial_1 u) + \partial_2 (\partial_2 u) & = \partial_1(3x^2 - 3y^2) + \partial_2(-6xy) = 6x - 6x = 0 \\ \partial_1 (\partial_1 v) + \partial_2 (\partial_2 v) & = \partial_1(3x^2 - 3y^2) + \partial_2(-6xy) = 6x - 6x = 0 \end{align*} as desired.
We have \begin{align*} (\partial_1 u)(x,y) & = 5x^4 - 30x^2 y^2 + 5y^4 \\ (\partial_2 u)(x,y) & = -20x^3 y + 20xy^3 \end{align*} so \begin{align*} (\partial_1 v)(x,y) & = 20x^3 y - 20 x y^3 \\ (\partial_2 v)(x,y) & = 5x^4 - 30x^2 y^2 + 5y^4 \end{align*} from the Cauchy-Riemann equations. Partially integrating gives \[ 5x^4 y - 10 x^2 y^3 + y^5 + h(x) = v(x,y) = 5 x^4 y - 10x^2 y^3 + g(y) \] which will be satisfied if e.g. $h(x) = 0$ and $g(y) = y^5$. Then \[ f(x+iy) = (x^5 - 10x^3 y^2 + 5xy^4) + i(5x^4 y - 10x^2 y^3 + y^5) = (x+iy)^5 \] and $f(z) = z^5$.
We have \begin{align*} (\partial_1 u)(x,y) & = 3x^2 - k y^2 + 12y - 12 \\ (\partial_2 u)(x,y) & = -2kxy + 12x \end{align*} so \begin{align*} (\partial_1 v)(x,y) & = 2kxy - 12x \\ (\partial_2 v)(x,y) & = 3x^2 - k y^2 + 12y - 12 \end{align*} from the Cauchy-Riemann equations. Partially integrating gives \[ 3x^2 y - \frac{k}{3} y^3 + 6y^2 - 12y + h(x) = v(x,y) = kx^2y - 6x^2 + g(y) \] and therefore \[ 3x^2 y = kx^2 y \] if $h(x) = -6x^2$ and $g(y) = -\frac{k}{3} y^3 + 6y^2 - 12y$. Therefore, the only value of $k$ for which $u$ could possibly be the real part of a holomorphic function is $3$. When $k = 3$ we have \[ v(x,y) = 3x^2 y - 6x^2 - y^3 + 6y^2 - 12y \] and $u,v$ are the real and imaginary parts of $f(z) = z^3 + 6 iz^2 - 12z$.
If $u$ is constant and the real part of a holomorphic function then the Cauchy-Riemann equations would force $(\partial_1 v)(x,y) = 0 = (\partial_2 v)(x,y)$ whence $v$ is constant as well.
In the purported situation the Cauchy-Riemann equations become \begin{gather*} u'(x) = (\partial_1 u)(x,y) = (\partial_2 v)(x,y) = v'(y) \\ 0 = (\partial_2 u)(x,y) = - (\partial_1 v)(x,y) = 0 \end{gather*} which tells us that $u'$ and $v'$ are constant. But then $u(x) = \mu x + b$ and $v(y) = \nu y + c$. Put $\lambda = \mu + i \nu$ and $a = b + ic$.
If we apply $\partial_1$ to the equation we get \[ 0 = 2(\partial_1 u)(x,y) + (\partial_1 v)(x,y) = 2(\partial_2 v)(x,y) + (\partial_1 v)(x,y) \] from the Cauchy-Riemann equations. We get \[ 0 = 2(\partial_2 u)(x,y) + (\partial_2 v)(x,y) = -2(\partial_1 v)(x,y) + (\partial_2 v)(x,y) \] if we apply instead $\partial_2$. Adding twice the first equation to the second gives $5 (\partial_2 v)(x,y) = 0$ for all $(x,y)$. It then follows from the second equation that $(\partial_1 v)(x,y) = 0$ for all $(x,y)$. The Cauchy-Riemann equations then gives $(\partial_1 u)(x,y) = 0$ and $(\partial_2 u)(x,y) = 0$ for all $(x,y)$. We conclude that $u$ and $v$ are constant, so that $f$ is as well.
Write $h = a + ib$. We need to verify that the limit \[ \lim_{h \to 0} \dfrac{f(h) - f(0)}{h} = \lim_{h \to 0} \dfrac{\sqrt{|ab|}}{h} \] does not exist. Taking $h = t$ with $t$ real gives the limit \[ \lim_{t \to 0} \dfrac{\sqrt{0}}{t} = 0 \] while taking $h = t+it$ with $t$ real gives the limit \[ \lim_{t \to 0} \dfrac{\sqrt{t^2}}{t+it} = \dfrac{1}{1+i} \] so the complex limit does not exist and $f$ is not differentiable at $0$.
We have $u(x,y) = \sqrt{|xy|}$ and $v(x,y) = 0$. The partial derivatives at the origin are \begin{align*} (\partial_1 u)(0,0) & = \lim_{t \to 0} \dfrac{u(t,0) - u(0,0)}{t} = \lim_{t \to 0} 0 = 0 \\ (\partial_2 u)(0,0) & = \lim_{t \to 0} \dfrac{u(0,t) - u(0,0)}{t} = \lim_{t \to 0} 0 = 0 \end{align*} and $(\partial_1 v)(x,y) = 0 = (\partial_2 v)(x,y)$. The Cauchy-Riemann equations are certainly satisfied at the origin. However, for example, the partial derivative $\partial_2 u$ is not defined on any ball centered at the origin, because \[ \lim_{t \to 0} \dfrac{u(a,t) - u(a,0)}{t} = \lim_{t \to 0} \dfrac{\sqrt{at}}{t} = \lim_{t \to 0} \dfrac{\sqrt{a}}{\sqrt{t}} \] does not exist and therefore $(\partial_2 u)(a,0)$ does not exist for any $a > 0$.