### Author Topic: LEC0101-Quiz7-ONE-E  (Read 638 times)

#### Xuefen luo

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##### LEC0101-Quiz7-ONE-E
« on: December 09, 2020, 12:21:13 PM »
Problem: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the first quadrant:
$f(z)=z^4-3z^2+3$

Answer: According to the picture, we divide the line into three parts, denoted $\gamma_1, \gamma_2, \gamma_3$.

For $\gamma_1$: $z=x, x\in [0,R]$ with $R \rightarrow \infty$.
$f(z) = x^4-3x^2+3$.
$f(z)$ is real number as $x$ traverses from $0$ to $R$.
Thus, change of argument of $f(z)$ is $0$.

For $\gamma_2$: $z=Re^{it}, t\in [o,\frac{\pi}{2}], R \rightarrow \infty$.
$f(z) = R^4e^{i4t}-3R^2e^{i2t}+3 =R^4(e^{i4t}-3\frac{R^2}{R^4}e^{i2t}+\frac{3}{R^4}) =R^4e^{i4t}$, as $R \rightarrow \infty$
Then, $4t\in [0,2\pi]$ and change of argument of $f(z)$ is $2\pi$.

For $\gamma_3$: $z=yi, y\in [R,0]$ with $R \rightarrow \infty$.
$f(z)= y^4+3y^2+3$
$f(z)$ is real number as $y$ traverses from $R$ to $0$.
Thus, change of argument of $f(z)$ is $0$.

Since $f(z)$ has no poles, we know the number of zeroes of $f(z)$ in the first quadrant is $\frac{1}{2\pi}(0+2\pi+0)=1$ by the argument principle.

#### Xuefen luo

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##### Re: LEC0101-Quiz7-ONE-E
« Reply #1 on: December 09, 2020, 01:25:05 PM »
Here is the given picture.

#### Xun Zheng

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##### Solved
« Reply #2 on: December 12, 2020, 04:24:57 PM »
Solved