Vivian

Welcome to Vivian's Page!

picture_of_log_mapping.png

Question 1

Evaluate the following functions at the indicated points:

a) $f(x) = z^{3}+2z$ at $z = i$

(1)
\begin{equation} f(i) = i^{3}+2i \end{equation}
(2)
\begin{equation} = -i + 2i \end{equation}
(3)
\begin{equation} = i \end{equation}

b) $f(z) = e^{z}$ at $z=2+\frac{\pi}{3}i$

(4)
\begin{align} f(i) = e^{2+\frac{\pi}{3}i} \end{align}
(5)
\begin{align} = e^{2}e^{\frac{\pi}{3}i} \end{align}
(6)
\begin{align} = e^{2}(\cos (\frac{\pi}{3})+i \sin (\frac{\pi}{3})) \end{align}
(7)
\begin{align} = e^{2}\cos (\frac{\pi}{3})+ie^{2} \sin (\frac{\pi}{3}) \end{align}

c) $f(z)= z+\frac{1}{z}$ at $z= 1 +2i$

(8)
\begin{align} 1 +2i + \frac{1}{1+2i} *\frac{(1-2i)}{(1-2i)} \end{align}
(9)
\begin{align} 1 +2i + \frac{1-2i}{1+4} \end{align}
(10)
\begin{align} 1 +2i + \frac{1-2i}{5} \end{align}
(11)
\begin{align} \frac{5+10i}{5}+\frac{1-2i}{5} \end{align}
(12)
\begin{align} \frac{6+8i}{5} \end{align}

Question 1

Use the Cauchy-Riemann equations to show that the following functions are analytic everywhere. Also find the derivative.

a) $f(z) = z^{3}$

(13)
\begin{equation} z = x+iy \end{equation}
(14)
\begin{equation} (x+iy)^{3} \end{equation}
(15)
\begin{equation} (x+iy)(x+iy)(x+iy) \end{equation}
(16)
\begin{equation} x^{3}+x^{2}iy+2x^{2}iy-2xy^{2}-xy^{2}-iy^{3} \end{equation}
(17)
\begin{equation} u = x^{3}-3xy^{2} \end{equation}
(18)
\begin{equation} v = 3x^{2}y - y^{3} \end{equation}
(19)
\begin{align} \frac{\partial u}{\partial x} = 3x^{2}-3y^{2}\;\;\; \frac{\partial v}{\partial y} = 3x^{2}-3y^{2} \end{align}
(20)
\begin{align} \frac{\partial v}{\partial x} = 6xy \;\;\; -\frac{\partial u}{\partial y} = 6xy \end{align}
(21)
\begin{align} \frac{\partial w}{\partial z}=\frac{\partial u}{\partial x} +\frac{i\partial v}{\partial x} \end{align}
(22)
\begin{align} \frac{\partial w}{\partial z}= 3x^{2}-3y^{2}+6xyi \end{align}
(23)
\begin{align} \frac{\partial w}{\partial z}= 3(x^{2}-y^{2}+2xyi) \end{align}
(24)
\begin{align} \frac{\partial w}{\partial z}= 3(x+iy)^{2} \end{align}
(25)
\begin{align} \frac{\partial w}{\partial z}= 3z^{2} \end{align}

Therefore the derivatives of $z^{3}$ is $3z^{3}$

Question 2

c. $\frac{1}{z(z-3)}$ $\;\;\;$ on annulus $0<|z-3|<3$ $\;\;\;\;\;$ expand in powers of $z-3$.

(26)
\begin{align} \frac{1}{z(z-3)} = \frac{1}{z-3} \left [\frac{1}{z} \right] \end{align}
(27)
\begin{align} \frac{1}{z-3} \left [ \frac{1}{3+(z-3)} \right ] = \frac{1}{z-3} \left [\frac{1}{3}\left ( \frac{1}{1+\frac{z-3}{3}} \right ) \right] \end{align}
(28)
\begin{align} \frac{1}{3(z-3)} \left [ 1+\frac{z-3}{3}+\frac{z-3}{3}^{2} +\frac{z-3}{3}^{3}+\frac{z-3}{3}^{4}+... \right ] \end{align}
(29)
\begin{align} \frac{1}{\, 3(z-3)}+ \frac{1}{9}+\frac{z-3}{27}+\frac{z-3^{2}}{81} +\frac{z-3^{3}}{243}+.... \end{align}

++Question 1

Compute the residue at the poles of each of the following functions.

a) $f(z)=\frac{1}{z^{2}-6z+8}$

(30)
\begin{align} \frac{1}{z^{2}-6z+8} \end{align}
(31)
\begin{align} \frac{1}{(z-4)(z-2)} \end{align}

Simple poles at $z=4$ and $z=2$.

(32)
\begin{align} Res(4)= \frac{1}{2z-6} = \frac{1}{2(4)-6} = \frac{1}{2} \end{align}
(33)
\begin{align} Res(2)= \frac{1}{2z-6} = \frac{1}{2(2)-6} = -1 \end{align}
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License