Solutions Dr. Z.'s Intro to Complex Variable Attendance Quiz for Lecture 24

1. Find the number of zeros of the function

f(z)=z^9-1-I

in the SECOND quadrant.

So. to 1: It just occurred to me, that for solving this problem,   you don't need to remember the acronym ASTC

(First Quadrant: A: ALL trig functions (cos,sin,tan) are Positive
 Second  Quadrant:  S:  sin is positive
 Third  Quadrant:  T:  tan is positive
 Fourth  Quadrant:  C:  cos is positive)

Since we are doing everything in POLAR notation z=r*exp(I*t)= r*(cos(t)+ I*sin(t)), all you have to remember is

First Quadrant: 0<t<Pi/2

Second Quadrant: Pi/2<t<Pi

Third Quadrant: Pi<t<3*Pi/2 (or -Pi<t< -Pi/2 , with the usual convention for the Argument)

Fourth Quadrant: 3*Pi/2<t<Pi (or -Pi/2<t< 0 , with the usual convention for the Argument).

We have to solve z^9=1+I

1+I=sqrt(2)*exp(I*Pi/4)

Writing z=r*exp(I*t), we get z^9=r^9*exp(9*I*t) so r^9=2^(1/2) and r=2^(1/18) (but this is irrelevant for the answer)
and

9*t=(Pi/4+2*n*Pi), 0<=n<9

So

t=(2*n+1/4)/9*Pi , 0<=n<9

We want Pi/2<t<Pi so

 1/2 <(2*n+1/4)/9<1

So

9/2<2*n+1/4< 9 

So

17/4<2*n<35/4

So

17/8<n<35/8

2+1/8<n<4+3/8

So among n=0,1,..., 8, only n=3, n=4 qualify and the two desired roots are
2^(1/8)* exp((2*n+1/4)/9*Pi*I), n=3,n=4 i.e.

2^(1/8)*exp((6+1/4)/9*Pi*I), 2^(1/8)*exp((8+1/4)/9*Pi*I),

Ans. to 1: The number of roots of f(z)=z^9-1-I in the second quadrant is 2.









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