Author Topic: LEC0101 Quiz#7 ONE-B  (Read 399 times)

Xun Zheng

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LEC0101 Quiz#7 ONE-B
« on: December 12, 2020, 04:26:52 PM »
Using argument principle along line on the picture, calculate the number of zeroes of the following function in the first quadrant:
$$f(z)=z^9+5z^2+3$$

Answer: First, we separate the line into three parts, denoted (1), (2), and (3).

For (1): $z=x, x: 0 \rightarrow R$ with $R \rightarrow \infty$.
$$f(z) = f(x) = x^9+5x^2+3$$
Thus we have $$arg(f(z))=0$$
Hence, the change of argument is $0$.


For (2): $z=Re^{it}, t: 0 \rightarrow \frac{\pi}{2}$ with $R \rightarrow \infty$.
$$f(z) = R^9e^{i9t}-5R^2e^{i2t}+3 =R^9(e^{i9t}-\frac {5e^{i2t}} {R^7}+\frac{3}{R^9})$$
As $R \rightarrow \infty$, $f(z) \rightarrow R^9e^{i9t}$, where $9t\in [0,\frac {9\pi} {2}]$.
Hence, the change of argument is $\frac {9\pi}{2}$.


For (3): $z=yi, y: R \rightarrow 0$ with $R \rightarrow \infty$.
$$f(z)= iy^9-5y^2+3$$
As $R \rightarrow \infty$, $f(z) \rightarrow iy^{9}$.
As $y = 0$, $f(z) = 3$.
Hence, $f(z)$ traverses from y-axis to $(3,0)$.
Thus, the change of argument is $$0 - \frac{\pi}{2} = \frac{-\pi}{2}$$


Therefore, by definition, $f(z)$ has 2 zeroes in the first quadrant.

Xun Zheng

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Re: LEC0101 Quiz#7 ONE-B
« Reply #1 on: December 12, 2020, 04:34:27 PM »
Here is the graph for this question