# 复杂函数 Complex Functions MATH2007

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$\quad w=\sqrt{z}$.
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
$$\begin{gathered} z=\cos 2 \pi+i \sin 2 \pi, \ w=\sqrt{z}=\cos \pi+i \sin \pi=-1 \end{gathered}$$

## 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

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$$\frac{d w}{d z}=\frac{\delta w}{8 x}=\frac{1}{i} \frac{\delta v}{\delta y}$$
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
$$\frac{\delta}{\delta y}\left(\frac{d u t}{d z}\right)=i \frac{\delta}{\delta x}\left(\frac{d u t}{d z}\right) \text {, }$$

## MATH243COURSE 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