# Toronto Math Forum

## MAT334--2020S => MAT334--Lectures & Home Assignments => Chapter 3 => Topic started by: Yan Zhou on March 15, 2020, 10:17:20 PM

Title: 3.1 question 5
Post by: Yan Zhou on March 15, 2020, 10:17:20 PM
$$f(z) = z^9 + 5z^2 + 3$$
I have difficulty in figuring out how f moves on $iy$.
The following is my steps.
$$f(iy) = iy^9 - 5y^2 + 3$$
y moves from R to 0. Im(f) > 0, f always lie in first or second quadrant.
f(0) = 3 on the real axis.
$$f(iR) = iR^9 - 5R^2 + 3$$
$$arg(f) = arctan(\frac{R^9}{-5R^2 + 3})$$
As R goes to $\infty$, arg(f) goes to $\frac{\pi}{2}$,
Then arg(f) changes from $\frac{\pi}{2}$ to 0, then $\Delta arg(f) = -\frac{\pi}{2}$
I am not sure about arctan part, should it goes to $\frac{\pi}{2}$ or  $-\frac{\pi}{2}$
and if there is any other mistakes, thanks for pointing out!