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.
