Prove: (z̄ )^k=(z̄ ^k) for z≠0 when k is negative

• shannon
In summary: So you can just move the * from the parenthesis to the outside. So (k*i)* = k*i*. Then you can just combine the exponents as usual to get (z^k)^*. In summary, to prove that (z̄)^k = (z̄^k) for every integer k (provided z≠0 when k is negative), write z in polar form and use the fact that complex conjugation means replacing "i" with "-i" to combine the exponents and get (z^k)^*.
shannon

Homework Statement

Prove that (z̄ )^k =(z̄ ^k) for every integer k (provided z≠0 when k is negative)

The Attempt at a Solution

I let z=a+bi so, z̄ =a-bi
Then I plugged that into one side of the equation to get
(a-bi)^k
I was going to try to manipulate this to get [(a-bi)^k]
But I don't know where to go from here, or even what to do...

shannon said:

Homework Statement

Prove that (z̄ )^k =(z̄ ^k) for every integer k (provided z≠0 when k is negative)
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

$$(z^*)^k = (z^{* k})$$ ??

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

$$(z^*)^k = (z^k)^*$$

was that the question?

If so, then write the complex number z in polar form $r e ^{i \theta}$

Is the bar over the z only in both parts? Or is it

$$\bar{z}^{k} = \bar{\left(z^k\right)}$$

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

$$(z^*)^k = (z^{* k})$$ ??

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

$$(z^*)^k = (z^k)^*$$

was that the question?

If so, then write the complex number z in polar form $r e ^{i \theta}$

Yes, that is my question. Sorry for the lack of clarity.
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).
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!

shannon said:
Yes, that is my question. Sorry for the lack of clarity.
No problem!
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).
Yes. Complex conjugation just means that we replace any "i" we see by -i. So yes, the sign of the exponent changes.
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!
Exactly! Basically, k times (i)* gives the same thing as (k times i)*

What is the statement being proven?

The statement being proven is that for any complex number z ≠ 0 and any negative integer k, (z̄ )^k=(z̄ ^k).

Why is the condition z ≠ 0 necessary?

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.

What does the notation z̄ represent?

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.

What is the significance of k being a negative integer?

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.

Can you provide an example to demonstrate this statement?

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.

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