Solutions to Dr. Z.'s Intro to Complex Variable Real Quiz #9

VERSION OF APRIL 24, 2020 (Thanks to Vishal Patel)

1. Use the Argument Principle (No credit for other methods) to find the number of zeros of

f(z)=z+i

(a) (3 points) In the  first quadrant {z: Re z>0, Im z>0}

Sol. to (1a). Consider the quarter circle x^2+y^2=R^2 in the first quadrant, where R is HUGE.
It has three segments

First Segment:
z=0 to z=R: at the beginning f(0)=i, so the argument is Pi/2. As z progresses along the x-axis in the
positive direction, f(z)=f(x)=x+i slides along the horizontal line y=1, ending at R+i, that is practically R, (pointing
rightward), and now the argument is 0, so

Net Gain of argument along first segment: 0-Pi/2= -Pi/2

Second Segment:
z=R to z=iR (along the circular arc). f(z)=R*exp(it)+i VERY close to R*exp(it), the argument is t, so as
t goes from 0 to Pi/2, the 

Net Gain of argument along second segment: Pi/2-0= Pi/2

Third Segment: From z=iR  back to z=0. Now z=iy (y>0) and f(z)=iy+i=(y+1)i, so the output, f(z) is ALWAYS
along the positive y-axis, always pointing up, in other words ALWAYS Pi/2, it does not change at all.

Net Gain of argument along third segment: Pi/2-Pi/2= 0

Adding up: The Change of argument when traveling along the quarter circle in the first quadrant with HUGE radius R is
-Pi/2+Pi/2+0=0

Dividing by 2*Pi, we get

Ans. to (1a): The number of zeros of f(z)=z+i in the first quadrant is 0.



(b) (3 points) In the second quadrant {z: Re z<0, Im z>0}

This is exactly the same way, but now the contour is the quarter circle x^2+y^2=R^2 in the second quadrant, where R is HUGE.


First  Segment: From z=0  back to z=iR. Now z=iy (y>0) and f(z)=iy+i=(y+1)i, so the output, f(z) is ALWAYS
along the positive y-axis, always pointing up, in other words ALWAYS Pi/2, it does not change at all.

Net Gain of argument along first segment: Pi/2-Pi/2= 0

Second Segment: z=iR to z=-R (along the circular arc). f(z)=R*exp(it)+i VERY close to R*exp(it), the argument is t, so as
t goes from Pi/2 to Pi, so

Net Gain of argument along second segment: Pi-Pi/2= Pi/2

Third segment: z=-R to z=0: at the beginning f(-R)=-R+i, practically -R, pointing to the left, the argument is
Pi-tiny (note that it is ABOVE the x-axis), so we use Pi, not -Pi.
As z progresses along the x-axis in the
positive direction, f(z)=f(x)=x+i slides along the horizontal line y=1, ending at i, with argument Pi/2
and now the argument is Pi/2, so

Net Gain of argument along third segment: Pi/2-Pi= -Pi/2 .

Adding up: 0+ Pi/2 + (-Pi/2)=0

Dividing by 2*Pi, we get 0/(2*Pi)=0

Ans. to (1b): The number of zeros of f(z)=z+i in the second quadrant is 0.


2. (2 points) Can you use the argument principle to find the number of zeros of f(z)=z+i in the third quadrant?
   In the fourth quadrant? EXPLAIN!

Sol. to 2: NO! The argument principle does not work if the function has poles or zeros on the contour. Since f(z) has a zero
at z=-i, and z=-i lies on the negative y-axis, that is part of the contour, the argument principle is
not applicable.

COMMENT: Of course we can consider a quarter-circle that excludes the point z=-i,  by  going parallel to the y-axis slightly to the
left and right (in the third and fourth quadrants, respectively), but this is a different story.