Complex Numbers: Solutions for z^n=a+bi , where |a+bi|= 1.

[Komandos]

Homework Statement

Hello everyone :) ok so that is a problem involving complex numbers and its a bit challenging, so i would be really glad if i could get some help with it!
The problem is:

Consider the complex equation
z^n=a+bi , where |a+bi|= 1.

I am supposed to generalize and prove results for this equation.
For simplicity, we can substitute for n=3,4,5 so that i can obtain a solution.
That would be the first part.

The second part is:
What happens if |a+bi|≠1

As i said, it might be a lil challenging but i hope anyone in this forum can help me!
Thank youuuu :)

Homework Equations

The Attempt at a Solution

 

[mfb]

Do you know how to express complex numbers with r and θ? Like $z=r e^{i \theta}$?
That simplifies the solution a lot.

 

[Komandos]

Wow thanks for the reply! :D
Soo you i do, but in school we're used to the polar form : z=rcisθ ... I have already found a lot of things previously .. cause this question is divided into many parts .. these are the last two ... so if you need additional information i can give it to you :)

 

[Komandos]

I have good news! I found the answer! :) But i need to prove it now :/ Hope someone will help mee!

Heres what i got:

z=(1/n)*(a/√(a²+b²)+ib/√(a²+b²))

(I think its right)

Thank you!

 

[mfb]

This would mean nz is constant - and |z|<1, which implies |z^n|<1, which is wrong.

 

[Komandos]

Aghh :/ ... ehm ok i have a different question.

How can i find the roots using polar form for:

z^n=i

 

[mfb]

Write i in polar form, and it should be obvious.

 

[Komandos]

Ok, so i have reached to some good results. However i still have a last question.

What we know is that the product of the distances of an complex number equation is equal to the number of the roots of it. So that we have:

if z^n=1 or if z^n = i
n=|1-w||1-w^2 |…|1-w^(n-1)|

That is a proof. Now the question is, is the same true for z^n = a+bi, where |a+bi| = 1.
The solution for this equation is:

z= cis((arctan(b/a)+2kπ)/n)

I have found out previously, that when solving z^n = i , the conjecture stated above is true with the only difference that the angle will change with an additional π /2. Since the modulus is in either way 1, i predict that the conjecture will be true for z^n = a+bi where |a+bi| = 1 , but i don't know how to prove it :/
So please , help! :(

Btw ... I really thank you for the answers and your time :)

 

[mfb]

You can try to follow the original proof, but take other numbers instead of i*. There are just two options:
- the proof does not work at some point
- the proof works

Alternatively, check it for some simple cases first.

*with z^n=1, z=1 is a solution, and |1-1|=0...

 

[HallsofIvy]

Wow thanks for the reply! :D
Soo you i do, but in school we're used to the polar form : z=rcisθ ... I have already found a lot of things previously .. cause this question is divided into many parts .. these are the last two ... so if you need additional information i can give it to you :)

Same thing. $e^{ix}= cos(x)+ i sin(x)$ so we can write polar form as $re^{i\theta}= r(cos(\theta)+ i sin(\theta))= r cis(\theta)$ where the last, $r cis(\theta)$, is just "engineer's shorthand" for $r(cos(\theta)+ i sin(\theta)$.

 

[Komandos]

Yaa thanks I know I said its the last question .. but apparaetly i wasnt done yet. So here's one more thing.

Lets consider the equation:
z^n= a+bi, where |a+bi| ≠ 1

Now if we want to find the general forumla for the roots we will have:

z= |a+bi|^1/n cis (∂+2kπ) , where ∂=arctan(b/a) (However we can leave ∂
as it is)

My question is how to find the general formula for the distances between the root where k=0 and the other roots of the equation.

Here we can use the general distance formula
√(x2-x1)²+(y2-y1)²

where , (x1,y1) = (0,0)

But however what should i put for (x2,y2) ?

 

[mfb]

I think there is a factor 1/n missing at "+2kπ".
You can use k=0, k=1 to find the distance between two adjacent solutions.

 

[Komandos]

ya great i did that. And you u were right :P 1/n was missing, I forgot to type it ... Anyways thanks for everything! Really it was such a great help! :)