Toronto Math Forum

Let f(z) = $2z^4 -2iz^3 +z^2 + 2iz -1$

Let z = $Re^{i\theta}$, and  $0\leq \theta \le \pi$  where  $R\to \infty$

$f(Re^{i\theta}) = 2(Re^{i\theta})^4 -2i(Re^{i\theta})^3 +(Re^{i\theta})^2 + 2i(Re^{i\theta}) -1$

$f(Re^{i\theta}) = R^4e^{4i\theta}(2- \frac{2i}{Re^{i\theta}} + \frac{1}{R^2e^{2i\theta}} +\frac{2}{R^3e^{3i\theta}} - \frac{1}{R^4e^{4i\theta}})$

$\triangle arg f(Re^{i\theta}) = 4\pi$    This was a standard part V.I.


On the x-axis:

if $z = x$
$$f(x) = 2x^4 -2ix^3 + x^2 + 2ix -1= 2x^4 +x^2 -1-2ix(x^2 -1)
$$
The real part is $\Re(f(x))=2x^4 +x^2 -1$, and the imaginary part is $\Im(f(x)) =2x(x^2 -1)$.

As for the imaginary part = 0, $x = \pm 1, 0$.

As for the real part = 0, $x =  \frac{-1\pm 3}{4}$. WRONG

If $x < -1, Re f >  0,  Imf > 0$, in the first quadrant.

If $-1 < x < \frac{-1-3}{4}, Ref > 0, Imf < 0$, in the fourth quadrant.

Therefore, $\triangle argf(x)\arrowvert_{\gamma x} = -2\pi$

$N - P = \frac{1}{2\pi}\triangle argf(z)\arrowvert_{\gamma} = \frac{\triangle argf(x)\arrowvert_{\gamma x} + \triangle argf(Re^{i\theta})\arrowvert_{\gamma R}}{2\pi} = 1$

The number of zero is 1.