For The Complex Numbers Find The Following

Find all complex numbers satisfying z^6 =1?

AndyJ has the answer if you want to find the 6 solutions of z by factoring. However, these problems generally use De Movire's Theorem to solve them.

We see that:

z^6 = 1
==> z = 1^(1/6).

So our objective is to find the 6 roots of a unity.

We know that:

1 = cos(0) + i sin(0).

Then, applying De Movire's Theorem:

1^(1/6) = cos[(0 + 2πk)/6] + i sin[(0 + 2πk)/6]
==> 1^(1/6) = cos(πk/3) + i sin(πk/3) for k = 0, 1, 2, 3, 4, 5.

Thus, the desired solutions are:

k = 0 ==> z = cos(0) + i sin(0) = 1
k = 1 ==> z = cos(π/3) + i sin(π/3) = (1/2) + (√3/2)i
k = 2 ==> z = cos(2π/3) + i sin(2π/3) = (-1/2) + (√3/2)i
k = 3 ==> z = cos(π) + i sin(π) = -1
k = 4 ==> z = cos(4π/3) + i sin(4π/3) = (-1/2) - (√3/2)i
k = 5 ==> z = cos(5π/3) + i sin(5π/3) = (1/2) - (√3/2)i.

I hope this helps!

Find all solutions in complex numbers z to the following equation (z-3)^4=2i?

The answer is z=3+ the four fourth roots of (2i)
So you have 4 possible answers.
To find the four possible fourth roots of 2i, we use de Moivre's theorem:
-First put 2i in polar form: 2e^(i*pi/2)
-The roots are: (fourth root of 2) *(pi/8 + 2n*pi/4), n=0,1,2,3
if we call the fourth root of 2 =a, then the four possible fourth roots of 2i are: a + pi/8, a+5*pi/8, a + 9*pi/8 and a+13*pi/8
Add 3 to each of the above values, and you get the possible values of z.

Find the quotient of the complex numbers?

Multiply by the Conjugate of the denominator, basically the denominator itself but with the opposite sign. This will become a sum/difference of squares and will cancel the middle terms when you foil and then you can substitute -1 in for i squared.

So, Multiply the top and bottom by (3+i) and then foil the top and bottom to get
24-18i+8i-6i^2 / 9-i^2 = 24-10i-6(-1) / 9-(-1) = 24+6-10i / 10 = 30-10i / 10 = 3-(1/3)i

Find the quotient of the complex numbers...?

(4 + 4i)/(5 + 4i)

You have to rationalize using the conjugate factor (5 - 4i)/(5 - 4i)

(4 + 4i)/(5 + 4i) * [(5 - 4i)/(5 - 4i)] = (20 + 4i +16)/41 =

(36 + 4i)/41

QED

Complex numbers. If z=cos(2theta)+isin(2theta), find the modulus and argument of 1-z in....?

x = theta

1 - z = (1 - cos2x) - i sin2x

4-step process to find argument
1. find tan_inv(imaginary/real)
2. identify quadrant of the complex number by looking at signs of real and imaginary part
3. identify quadrant of step1
4. if quadrants in step2 and step3 do not match, add pi to answer of step1

i will only solve part a completely

1. as you've calculated, we have to find tan_inv(-cotx) = -tan_inv(cotx) = -tan_inv(tan(pi/2 - x)) = x - pi/2
2. re(1-z) = 1 - cos2x >= 0 : 1st or 4th quadrant
im(1-z) = -2sin2x which is
i) 0 ii) pi/23. i) 0 ii) pi/24. i) 0 ii) pi/2

in part b, i'm guessing the quadrants will not match, so answer will be x-pi/2+pi = x+pi/2







i'll put a second method

express complex number as r(cos y + i sin y), where r > 0
y is the argument

to use this method,
1 - z = (1 - cos 2x) - i sin 2x
= (2sinx*sinx) - i (2sinx*cosx)
= 2sinx(sinx - i cosx)
= 2sinx(cos(pi/2-x) - isin(pi/2-x))
= 2sinx(cos(x-pi/2) + isin(x-pi/2)) (cos(-x) = cosx, sin(-x) = -sinx)

for part a, sinx > 0, so this representation works
which gives arg = x - pi/2

for part b, sinx < 0, so to keep r positive, we must change to
1 - z = (-2sinx)(-cos(x-pi/2) - isin(x-pi/2))
= (-2sinx)(cos(x+pi/2) + isin(x+pi/2)) (sin(x + pi) = -sinx, cos(x+pi) = -cosx)
now, -2sinx is positive, so it is of the form r (cos y + isin y)
this gives arg = x + pi/2