2 - 2 powers and roots of complex no

Question 1

de moivres theorum z^n =

Answer 1

z^n = (r(cosθ + i sinθ))^n = r^n(cos nθ + i sin nθ)

Question 2

how to evaluate complex numbers to large powers

Answer 2

put in mod arg form
use demoivres theorem
convert back to cartesian

Question 3

writing complex numbers as powers of e

Answer 3

r(cos θ + i sinθ) = re^iθ

Question 4

complex conjugate in exponential form

Answer 4

z = r e^iθ
z* = r e^-iθ

Question 5

finding roots of an equation if z^n = x + iy

Answer 5

write x+ iy in mod arg form
use demoires theorum to find z^n
then compare the mod and arg to x + iy
find the other roots by finding the next angles + 2pi/n

Question 6

roots of z^n = w geometircally

Answer 6

will create a regular polygon with n vertices on a circel centred at 0,0 with radius = r
eg n = 3 is a triangle

Question 7

roots of unity - solving z^n = 1

Answer 7

1, e^2pii/n, e^4pii/n … e^2(n-1)pii/n

Question 8

nth root of unity

Answer 8

wk = (e^2pii/n)^k= w1^k

Question 9

sum of roots of unity

Answer 9

= 0
1 + w + w^2 ..w^n-1 (geometric series)
= 1-w^n/1-w = 0 as w^n = 1

Question 10

(z-w)(z-w*)

Answer 10

= z^2 - 2zRe(w) + w^2

Question 11

write z^n + c as 2 quadratic factors

Answer 11

eg z^4 + 81
solve z^4 = -81
then you have 4 factors - 2 pairs of conjugates
which you can simplify to z^2 - 2zRe(w) + w^2 twice

Question 12

multiplying by r cosθ + i sinθ as a tranformation

Answer 12

rotation by θ and enlargemnt by r
dividing is rotation of -θ and enlargemtn 1/r