# Thread: IS F(Z) = Z6^ analytic

1. ## IS F(Z) = Z6^ analytic

Hi guys, would be very grateful if someone could post a solution to this question

"Check if F(Z) = Z^6 is analytic"

2. One way:

$\displaystyle f'(z)=\displaystyle\lim_{h\to 0}\dfrac{(z+h)^6-z^6}{h}=\ldots=6z^5$

That is, for every $\displaystyle z\in\mathbb{C}$ there exists $\displaystyle f'(z)$ (finite) .

3. Thank You, just wondering if thats the full solution or do i continue getting the limit, i am unsure how to continue getting the limit.

4. Originally Posted by davy337
Thank You, just wondering if thats the full solution or do i continue getting the limit, i am unsure how to continue getting the limit.

Use the Newton's Bynomial Theorem, $\displaystyle (z+h)^6=z^6+6z^5h+\ldots+h^6$ , cancel $\displaystyle z^6$ and divide numarator and denominator by $\displaystyle h$ .

5. I feel terrible asking but would it be possibe to get the full solution from you ? apoligies for taking up your time

6. Originally Posted by davy337
I feel terrible asking but would it be possibe to get the full solution from you ? apoligies for taking up your time

I wouldn't mind, but is not good for you. Better study the corresponding theory and ask your doubts.

7. Originally Posted by FernandoRevilla
I wouldn't mind, but is not good for you. Better study the corresponding theory and ask your doubts.
I would like to support the above advice and remind members of MHF policy (clearly stated in the forum rules).