The discussion revolves around proving the equality \((\bar{z})^k = (\bar{z^k})\) for complex numbers, specifically when \(z \neq 0\) and \(k\) is negative. The context involves complex conjugation and properties of exponents in complex numbers.
Participants are actively questioning the clarity of the original problem statement and are discussing the correct interpretation of the notation. Some have provided guidance on expressing complex numbers in polar form and the effect of conjugation on the exponent.
There is uncertainty regarding the notation used for complex conjugation, with participants seeking clarification on whether the original statement was correctly interpreted. The discussion also touches on the implications of negative exponents in the context of complex numbers.
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!