Complex Numbers: nth Root Proof

[Nick_273]

Homework Statement

Really new to complex numbers so please forgive my ignorance. Prove that if the complex number z is an n^th root of the real number x then the complex conjugate z (z with horizontal line on top) is also an n^th root of x.

Homework Equations

i²=-1

The Attempt at a Solution

ⁿsqr(z) = x
z = xⁿ

also,
abs(z) = xⁿ

 

[Dick]

Hmm. i is a square root (i.e. 2nd root) of -1. |i|=1. 1 isn't a square root of -1. Did you mean x is a nonnegative real number?

 

[Nick_273]

Sorry, i meant z repeated, not absolute value... its late and I'm tired :P

Hmm. i is a square root (i.e. 2nd root) of -1. |i|=1. 1 isn't a square root of -1. Did you mean x is a nonnegative real number?

 

[Dick]

Sorry, i meant z repeated, not absolute value... its late and I'm tired :P

Ooohhh. That's not z repeated. The bar on the top notation means "complex conjugate". It's not the same thing as in 'repeated' in decimal expansions. I think you might want to look that up and figure out what it is and then we can continue this. In the meantime get some sleep. For example, how is x related to the complex conjugate of x is x is real?

 

[Nick_273]

Haha, OK. I guess I need to get the question right before i can even attempt a solution :P.

I'll post again once I make some progress,

Thanks for clearing that up.

 

[angus.myers]

Did you ever manage to get this problem solved?

z = a - ib and z= a+ib are complex conjugates are they not? Where both a and b are real numbers.

so in this case, if z = n(sqrt)ia

then the complex conjugate is = n(sqrt)-ia

is this correct?

 

[Dick]

Did you ever manage to get this problem solved?

z = a - ib and z= a+ib are complex conjugates are they not? Where both a and b are real numbers.

so in this case, if z = n(sqrt)ia

then the complex conjugate is = n(sqrt)-ia

is this correct?

If x is real and x=z^n, can't you think of a simple reason why the complex conjugate of z raised to the power n is also x? What rule would tell you z=sqrt(n)ia? That's wrong.

 

[angus.myers]

I think here it's more a case of what the complex conjugate of z actually is. I think that is where I am having difficulties! I've never touched on complex numbers before, I'm trying to read up on it but am obviously having basic ground work issues.

How would one go about finding the complex conjugate. Is there a set of guidelines for this?

 

[Dick]

I think here it's more a case of what the complex conjugate of z actually is. I think that is where I am having difficulties! I've never touched on complex numbers before, I'm trying to read up on it but am obviously having basic ground work issues.

How would one go about finding the complex conjugate. Is there a set of guidelines for this?

You already said what the complex conjugate is. If z=a+bi with a and b real, then the complex conjugate is a-bi. The fact you need to solve this is that the complex conjugate of (a+bi)^n is (a-bi)^n. I.e. $$\bar {z^n}={\bar z}^n$$. Do you know that?

 

[angus.myers]

Wow! I'm an idiot! haha, Thank you for that! I didn't think of it that way and I think I was trying to over complicate things! so
z = (a +ib)^n and z(bar) = (a - ib)^n

when asked:
Prove that if the complex number z is an nth root of the real number x then
z is also an nth root of x.

means that ib = 0, as i isn't a real number? so if x is a real number, ib holds a value that is equal to 0.

Therefore z = z(bar) when b = 0? Thank you for your help!

 

[Dick]

Wow! I'm an idiot! haha, Thank you for that! I didn't think of it that way and I think I was trying to over complicate things!


so
z = (a +ib)^n and z(bar) = (a - ib)^n

when asked:
Prove that if the complex number z is an nth root of the real number x then
z is also an nth root of x.

means that ib = 0, as i isn't a real number? so if x is a real number, ib holds a value that is equal to 0.

Therefore z = z(bar) when b = 0?


Thank you for your help!

You stated that in the most confusing way possible, but yes, I think you basically understand it. x=z^n. If you take the complex conjugate of both sides, the complex conjugate of x is x since it's real. And you know what happens to the other side. I hope as time goes on your explanations will become more clear.

 

[angus.myers]

Haha awesome! Thank you very much for your help! Much appreciated!