MAT334-2018F > Quiz-7

Q7 TUT 5201

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Muyao Chen:
$f(z) = z^{4} -3z^{2} + 3$ substitute $w = z^{2}$
$$z^{4} -3z^{2} + 3 = w^{2} - 3w + 3 = 0$$

$w = \frac{3 \pm i \sqrt 3}{2}$, so $z=\sqrt{\frac{3 \pm i \sqrt 3}{2}}$

For z in $[0, R]$:
$$f(x)= x^{4}-3x^{2}+ 3$$
$f(0) = 3$, then $arg(f(z)) = 0$

For $z = Re^{it}$ And $0 \leq t \leq \frac{\pi}{2}$:
$$f(Re^{it}) = R^{4}e^{4it} - eRe^{2it} + 3 = R^{4}(e^{4it} - \frac{3e^{2it}}{R^{3}} + \frac{3}{R^{4}}) = R^{4}e^{4it}(1 - \frac{3}{R^{3} e^{2it}} +\frac{3}{R^{4}})$$
As $R \rightarrow \infty$
$f(z)= R^{4}e^{4it}$ along 2, so that t goes from 0 to $\frac{\pi}{2}$, then $arg(f(z))$ goes from $4 *0 = 0$ to $4* \frac{\pi}{2}$ = 2$\pi$

For $z = iy$, And $0 \leq y \leq R$:
$$f(iy) = y^{4} - 3y + 3$$
$$f(0) = 3$$
Then $arg(f(z)) = 0$
Then $$\triangle arg(f(z)) = 0 +2 \pi + 0 = 2 \pi$$
$$\frac{1}{2 \pi}[\triangle arg(f(z))] = N_{0} - N_{p} = N_{0} = \frac{1}{2 \pi}2 \pi = 1$$

Then only one solution of the function in the first quadrant.

Victor Ivrii:
All solutions are wrong even if some answers are correct: When saying $\arg(f(+\infty)) =0$ and $f(0)=0$ you forgot that $\arg$ is defined modulo $2\pi n$. You must trace change of $\arg$ carefully

Zechen Wang:
Since in [0, positive infinity) f(x) is always positive and always stays on the real axis, the change in arg on [0, R] should be zero.

Victor Ivrii:

--- Quote from: Zechen Wang on December 01, 2018, 02:22:06 PM ---Since in [0, positive infinity) f(x) is always positive and always stays on the real axis, the change in arg on [0, R] should be zero.

--- End quote ---
It is a good start. What about $[Ri,0]$? And about arc (well someone wrote about arc, but I want everything in one place)

Zechen Wang:
here is the solution. since there are no changes along both positive axis, change of arg is zero on those lines. The only change of arg is on the arc = 2pi.