Polynomials Question (1 Viewer)

[cutemouse]

Hi,

I have problems with the following questions. Could you please show me how to do this, rather than just giving me the answer?

Find and plot on complex plane (I can plot myself )

1. z⁴+1=0

2. z²=i

3. z⁴+16=0

I tried doing these by making them into rcisθ, but I didn't get the answer.... Why doesn't this work?

Thanks,

Jason

 

[anon09]

This question should work if you switch the complex numbers to polar form.

This is what I get:

z4 = -1

z = rcis theta, therefore, z4 = r4cis 4theta (De Moivre's Theorem)

-1 = cis pi

Equate the two:

r4 cis 4theta = cis pi

r4 = 1

r = 1

and

4theta = pi + 2pi.k (accounting for revolutions)

theta = [pi (1 + 2k)] /4

Therefore:

The first root = cis pi/4 (when k = 0)

The second root = cis 3pi/4 (k=1)

The third root = cis 5pi/4 (k=2)

The fourth root = cis 7pi/4 (k=3)


Just follow the same procedure for the rest i.e

r2cis 2theta = cis pi/2

and

r4cis 4theta = 16cis pi

Good luck!

 

[tommykins]

z^4+1 = 0
z^4 = -1 = cis pi
z = cis[(pi+2kpi)/4] where k = -1,0,1,2

z^2 = i = cis pi/2
z = cis[(pi+2kpi/4)] wher k = 0, 1

same method for 3.

 

[youngminii]

z^4+1 = 0
z^4 = -1 = cis pi
z = cis[(pi+2kpi)/4] where k = -1,0,1,2

z^2 = i = cis pi/2
z = cis[(pi+2kpi/4)] wher k = 0, 1

same method for 3.

That's wrong
You forgot, z^2 = {cis[(pi + 4kpi)/2]}
z = cis[(pi+4kpi)/4] where k = 0, 1

I think post-HSC is killing your maths, tommy =D

 

[tommykins]

Yer probably.