Home | Assessment | Notes | Index | Worksheets | Blackboard
We multiple the right-hand side by $A-B$ to get \begin{align*} & A (A^{n-1} + A^{n-2} B^1 + \cdots + A^1 B^{n-2} + B^{n-1}) \\ & \qquad - B(A^{n-1} + A^{n-2} B^1 + \cdots + A^1 B^{n-2} + B^{n-1}) \\ = & (A^n + A^{n-1} B^1 + \cdots + A^2 B^{n-2} + A^1 B^{n-1}) \\ & \qquad - (A^{n-1} B^1 + A^{n-2} B^2 + \cdots + A^1 B^{n-1} + B^n) \\ = & A^n - B^n \end{align*} as desired.
We can differentiate the power series for $1 / (1-z)$ inside $\ball(0,1)$ term-by-term. This gives \[ \frac{1}{(1-z)^2} = 1 + 2z + 3z^2 + 4z^3 + \cdots \]
Replacing $z$ with $z^2$ in the geometric series gives \[ \frac{1}{1-z^2} = 1 + z^2 + z^4 + z^6 + \cdots \] on $\ball(0,1)$ because $|z|^2 = |z| |z| < 1$ whenever $|z| < 1$.
Differentiating our answer to (b) on $\ball(0,1)$ gives \[ \dfrac{2z}{(1-z^2)^2} = 2z + 4z^3 + 6z^5 + \cdots \] for all $|z| < 1$. \qedhere
Fix $w = u+iv \ne 0$. We need to produce $z$ with $\exp(z) = w$. From \[ \exp(x+iy) = e^x( \cos(y) + i \sin(y) ) \] we should take $x = \ln(|w|)$ and $y = \Arg(w)$.
Fix $z \in \C$. We calculate \begin{align*} \exp(z + 2 \pi i n) & = \exp(x + i (y + 2 \pi n)) \\ & = e^x (\cos(y + 2\pi n) + i \sin(y + 2\pi n)) \\ & = e^x(\cos(y) + i \sin(y)) \\ & = \exp(z) \end{align*} because $\sin$ and $\cos$ have $2\pi n$ as a period for any $n \in \Z$.
Suppose $\tau$ is a period of $\exp$. Then $\exp(z + \tau) = \exp(z)$ for all $z \in \C$. In particular \[ \exp(z) = \exp(z + \tau) = \exp(z) \exp(\tau) \] so $\exp(\tau) = 1$. Write $\tau = a + ib$. Then \[ 1 = e^{a} (\cos(b) + i \sin(b)) \] which forces $|1| = e^a$, $\cos(b) = 1$ and $\sin(b) = 0$. We must have $a = 0$ and $b = 2\pi n$ so $\exp$ has no other periods.
Write \[ \exp(x+iy) = e^x(\cos(y) + i \sin(y)) = e^x \cos(y) + i e^x \sin(y) \] so $u(x,y) = e^x \cos(y)$ and $v(x,y) = e^x \sin(y)$. Then \begin{align*} (\partial_1 u)(x,y) = e^x \cos(y) & \qquad (\partial_2 u)(x,y) = - e^x \sin(y) \\ (\partial_1 v)(x,y) = e^x \sin(y) & \qquad (\partial_2 v)(x,y) = e^x \cos(y) \end{align*} and the Cauchy-Riemann equations are seen to be true for all $(x,y)$.
If $\exp(x+iy)$ is real then $e^x \sin(y) = 0$ which is the same as $y = 2 \pi i n$ for some $n \in \Z$. If it is imaginary then $e^x \cos(y) = 0$ which is the same as $y = 2\pi i n + \frac{\pi}{2}$.
If $\exp(z) = -1$ then $e^x \cos(y) = -1$ and $e^x \sin(y) = 0$ so $x = 0$ and $y = 2\pi n + \pi$ for some $n \in \Z$.
If $\exp(z) = 1+i$ then $e^x \cos(y) = 1$ and $e^x \sin(y) = 1$. We have $e^x = |1 + i| = \sqrt{2}$ so $x = \frac{1}{2} \ln(2)$. From $\sin(y) = \frac{1}{\sqrt{2}} = \cos(y)$ we get $y = 2 \pi n + \frac{\pi}{4}$.
\[ \cos(i) = \dfrac{\exp(i^2) + \exp(-i^2)}{2} = \dfrac{\frac{1}{e} + e}{2} = \dfrac{1 + e^2}{2e} \] \[ \sin(i) = \dfrac{\exp(i^2) - \exp(-i^2}{2i} = \dfrac{\frac{1}{e} - e}{2i} = \dfrac{1 - e^2}{2ei} \]
No because the odd (respectively even) coefficients are zero: the quotient $a_{n+1} / a_n$ is not defined for all $n \in \N$.
Calculate \[ \begin{aligned} & \cos(z) \cos(w) - \sin(z) \sin(w) \\ = & \dfrac{\exp(iz) + \exp(-iz)}{2} \cdot \dfrac{\exp(iw) + \exp(-iw)}{2} \\ & \qquad - \dfrac{\exp(iz) - \exp(-iz)}{2i} \cdot \dfrac{\exp(iw) - \exp(-iw)}{2i} \\ = & \dfrac{\exp(i(z+w)) + \exp(i(z-w)) + \exp(i(-z+w)) + \exp(i(-z-w))}{4} \\ & \qquad + \dfrac{\exp(i(z+w)) - \exp(i(z-w)) - \exp(i(-z+w)) + \exp(i(-z-w))}{4} \\ = & \dfrac{\exp(i(z+w)) + \exp(-i(z+w))}{2} = \cos(z+w) \end{aligned} \] and \[ \begin{aligned} & \sin(z) \cos(w) + \sin(w) \cos(z) \\ = & \dfrac{\exp(iz) - \exp(-iz)}{2i} \cdot \dfrac{\exp(iw) + \exp(-iw)}{2} \\ & \qquad + \dfrac{\exp(iw) - \exp(-iw)}{2i} \cdot \dfrac{\exp(iz) + \exp(-iz)}{2} \\ = & \dfrac{\exp(i(z+w)) + \exp(i(z-w)) - \exp(i(-z+w)) - \exp(i(-z-w))}{4i} \\ & \qquad + \dfrac{\exp(i(w+z)) + \exp(i(w-z)) - \exp(i(-w+z)) - \exp(i(-w-z))}{4i} \\ = & \dfrac{\exp(i(w+z)) - \exp(i(-w-z))}{2i} = \sin(w+z) \end{aligned} \]
Calculate \begin{align*} \cosh(z) + \sinh(z) & = \dfrac{\exp(z) + \exp(-z)}{2} + \dfrac{\exp(z) - \exp(-z)}{2} \\ & = \exp(z) \end{align*}
Calculate \begin{align*} & (\cosh(z))^2 - (\sinh(z)^2 \\ = & \left( \dfrac{\exp(z) + \exp(-z)}{2} \right)^2 - \left( \dfrac{\exp(z) - \exp(-z)}{2} \right)^2 \\ = & \dfrac{\exp(2z) + 2 + \exp(-2z)}{4} - \dfrac{\exp(2z) - 2 + \exp(-2z)}{4} = 1 \end{align*}
We calculate $\gamma'(t) = 3 + i$ so that $|\gamma'(t)| = \sqrt{10}$. Then \[ \ell(\gamma) = \int\limits_3^5 \sqrt{10} \intd t = 2 \sqrt{10} \]
The following parameterization works. \begin{align*} \gamma_1(t) = 2t \textup{ on } [0,1] \\ \gamma_2(t) = 2 + 3ti \textup{ on } [0,1] \\ \gamma_3(t) = 3i + 2(1-t) \textup{ on } [0,1] \\ \gamma_4(t) = 3i(1-t) \textup{ on } [0,1] \end{align*}
A straight line from $i$ to $2$.
The reverse of $\Gamma = (\gamma_1,\gamma_2,\gamma_3)$ should be $(\tilde{\gamma_3},\tilde{\gamma_2},\tilde{\gamma_1})$.