Question

Find all the roots of the equation (2z-1)^3=1 in the form x+iy.

Answer 1

I have done it until here. How do I solve it further?
\[2\sqrt{2}z ^{3}-6z ^{2}+3\sqrt{2}z-2=0\]

Answer 2

The path you choose to find the solution seems a little hard. I would do something different. Try this,
\[(2z-1)^3=1\Rightarrow 2z-1=\theta\] where \theta is the 3rd rooth of 1. So there will be three equations,
\[ 2z-1=\theta_1\\
2z-1=\theta_2\\
2z-1=\theta_3\] And then solve for z.

Answer 3

do you know the cube roots of unity?

Answer 4

No, I don't know the cube roots of unity.

Answer 5

by using the polar form of z, you can find all z for which z^3=1....

Answer 6

Have you ever heard of De Moivre's Theorem?

Answer 7

then for those solutions, do what john_es said.

Answer 8

Yes. I know DeMoivre's Theorem.

Answer 9

It states that \[(\cos \theta+i \sin \theta)^n=\cos n \theta+i \sin n \theta\]

Answer 10

Okay, but how do I apply it the nth roots of unity?

Answer 11

Do you agree that \[\cos(\theta+2n \pi)=\cos \theta\]and\[\sin(\theta+2n \pi)=\sin \theta\] where n=1,2,3,...

Answer 12

correction: n=0,1,2,3,...

Answer 13

right?

Answer 14

Yes.

Answer 15

Substitute this into the original De Moivre's Theorem.

Answer 16

Ignore what I just said.

Answer 17

For an equation that looks like x^n=c where x is complex number and c is complex constant, we can apply de moivre's theorem.

Answer 18

I'm not following you. Can you please do it for one of the equations john posted? :o

Answer 19

sorry.
we can write c as following.\[c=a+b i=r(\cos \theta+i \sin \theta)\] where \[r=\sqrt{a^2+b^2}\]and\[\tan \theta=\frac{ b }{ a }\]

Answer 20

From the trig identities...\[c=r(\cos( \theta+2k \pi)+i \sin(\theta+2k \pi))\]

Answer 21

So, \[x^n=r(\cos(\theta+ 2k \pi)+i \sin(\theta+ 2k \pi))\]
Take the nth root to get\[x=r^\frac{ 1 }{ n }(\cos\frac{ \theta+2k \pi }{ n }+i \sin\frac{ \theta+2k \pi }{ n })\] where k=0,1,2,...,n-1

Answer 22

There, you have the general solution for x^n=c

Answer 23

The reason why k=0,1,2,..,n-1 is to create 'n' solutions.

Answer 24

May be, @emcrazy14 knows what is the polar form of a complex number. Then, 1 should be put in polar form first.
\[1=1_0\] And then do the root,
\[\sqrt[n]{a_\theta}=\sqrt[n]{a}_{\theta/n+2\pi k/n}\ \ k=0,1,n-1\] But it is the same as @science0229 said.

Answer 25

Right.

Answer 26

Okay, so I equate 2z-1 with the first root of unity, then the second and finally the third?

Answer 27

Yeah. Basically.

Answer 28

Also, for future reference, it would be good to memorize the "complex root formula"

Answer 29

Okay, thank you so much. Actually I didn't do this in class and we were taught the long way of forming the equation, then factorising it and equating to 0 to find the roots.

Answer 30

I forgot: make sure that the theta is between 0 and 2pi

Answer 31

And good luck!