复杂函数 Complex Functions MATH2007

0

这是一份nottingham诺丁汉大学MATH2007作业代写的成功案例

复杂函数 Complex Functions MATH2007

$\quad w=\sqrt{z}$.
In this $z=0$ is a branch-point. If we let the variable start from the point $z=1$ and describe the eireumference of a circle round the origin, this is a closed line which encloses the branch-point. If the fanction $w=\sqrt{z}$ start from the point $z=1$ with the value $w=+1$, and if we put
$$
z=r(\cos \phi+i \sin \phi),
$$
then at the point $z=1, r=1$ and $\phi=0$. If $z$ next describe the circumference of the circle in the direction of increasing angles, $r$ remains constant and equal to 1 , and $\phi$ inereases from 0 to $2 \pi$. If therefore the variable return to the point $z=1$, then
and therefore
$$
\begin{gathered}
z=\cos 2 \pi+i \sin 2 \pi, \
w=\sqrt{z}=\cos \pi+i \sin \pi=-1
\end{gathered}
$$

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MATH2007 COURSE NOTES :

$$
\sqrt[3]{\frac{z-a}{z-d} \cdot \frac{z-b}{z-e} \cdot \frac{(z-f)^{2}}{z^{2}}}
$$
by putting
$$
d=e=f=c .
$$
The branch-cuts can here also be chosen in a third way by drawing one from $a$ to 0 , and another from 0 to $b$.
The function
changes into
$$
\begin{aligned}
f(z) &=\sqrt[3]{(z-a)(z-b)} \
\phi(v) &=\sqrt[3]{\frac{(1-a u)(1-b u)}{u^{2}}}
\end{aligned}
$$








复杂函数 Complex Functions MATH243

0

这是一份liverpool利物浦大学MATH243的成功案例

复杂函数 Complex Functions MATH243

$$
\frac{d w}{d z}=\frac{\delta w}{8 x}=\frac{1}{i} \frac{\delta v}{\delta y}
$$
follow
und
$$
\frac{\delta}{\delta c}\left(\frac{d w}{d z}\right)=\frac{1}{i} \frac{\delta^{2} w}{\delta x \delta y}
$$
$$
\frac{\delta}{\delta y}\left(\frac{d v}{d z}\right)=\frac{\partial^{2} v}{8 x \delta y}
$$
sonsequently
$$
\frac{\delta}{\delta y}\left(\frac{d u t}{d z}\right)=i \frac{\delta}{\delta x}\left(\frac{d u t}{d z}\right) \text {, }
$$

英国论文代写Viking Essay为您提供作业代写代考服务

MATH243 COURSE NOTES :

$$
w^{3}-w+z=0
$$
If, for brevity, we put
$$
p=\sqrt[3]{\frac{1}{2}\left(-z-\sqrt{z^{2}-\frac{4}{27}}\right)}, q=\sqrt[3]{\frac{1}{2}\left(-z+\sqrt{\left.z^{2}-\frac{4}{27}\right)}\right.},
$$
and the two imaginary cube roots of unity
$$
\frac{-1+i \sqrt{3}}{2}=\alpha, \frac{-1-i \sqrt{3}}{2}=\alpha^{2}
$$
Cardan’s formula gives for the three roots of the above