This is a bit unclear. From the rest of your post, it looks as if by ̄ you meant the complex conjugation. So did you mean to writeshannon said:Homework Statement
Prove that (z̄ )^k =(z̄ ^k) for every integer k (provided z≠0 when k is negative)
nrqed said:This is a bit unclear. From the rest of your post, it looks as if by ̄ you meant the complex conjugation. So did you mean to write
[tex] (z^*)^k = (z^{* k}) [/tex] ??
But even this is unclear since most people would interpret the two sides to mean the same thing.
It would make more sense if the question was to prove
[tex] (z^*)^k = (z^k)^* [/tex]
was that the question?
If so, then write the complex number z in polar form [itex] r e ^{i \theta} [/itex]
No problem!shannon said:Yes, that is my question. Sorry for the lack of clarity.
Yes. Complex conjugation just means that we replace any "i" we see by -i. So yes, the sign of the exponent changes.So I rewrote z in polar form, but I wasn't sure about what the conjugate of that would be...would the exponent just be negated? (the iѲ term).
Exactly! Basically, k times (i)* gives the same thing as (k times i)*If that was the case, would I simply combine the exponents in the left side of the equation, multiplying the k and the iѲ term thus resulting in the right side of the equation?
I hope I'm being more clear!
The statement being proven is that for any complex number z ≠ 0 and any negative integer k, (z̄ )^k=(z̄ ^k).
The condition z ≠ 0 is necessary because the complex conjugate of 0 is still 0, so the statement would not hold true if z equals 0.
The notation z̄ represents the complex conjugate of the complex number z. This means that the imaginary part of z̄ is equal to the negative of the imaginary part of z.
The significance of k being a negative integer is that it allows us to generalize the statement for all values of k. If k were a positive integer, the statement would only hold true for specific values of z.
Yes, for example, let z = 2 + 3i and k = -2.Then, (z̄ )^k = (2 - 3i)^-2 = (1/13) - (6/13)iAnd (z̄ ^k) = (2 - 3i)^-2 = (1/13) - (6/13)iTherefore, (z̄ )^k = (z̄ ^k) for z ≠ 0 and k = -2.