Math Is Fun Forum

Daniel123

Member

Registered: 2007-05-23

Posts: 663

nth roots

The techniques used to find the nth roots of unity can be extended to find the nth roots of any complex number.

Suppose that

is an nth root of

then

The book then says that

and

, but I don't see why this would necessarily be the case?

Wouldn't you have to say that

and

?

Thanks.

Last edited by Daniel123 (2008-11-22 03:11:59)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: nth roots

surely you can see that (r')^n = r, and hence r' = n'th root of r ?
in which case, you can sub one for the other, and then just divide to get the result the book gives?

Last edited by luca-deltodesco (2008-11-22 12:56:23)

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Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: nth roots

"surely you can see that (r')^n = r"

No, I can't. Why does (r')^n = r? I would have thought all we can say is that:

and

.

Why can we assume that when taking the nth root of a number, the modulus of the nth root is equal to the nth root of the original modulus?

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: nth roots

so if A is the n'th root of B, it's modulus must be the n'th root of B's modulus, otherwise when you take A to the power of n, the modulus wouldn't equal that of B

Last edited by luca-deltodesco (2008-11-23 01:15:08)

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The Beginning Of All Things To End.
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Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: nth roots

Hmmm... I've been stupid. I can see it now.

Thanks