how to solve this small question?

[logistic_guy]

here is the question

Solve \(\displaystyle \int_{-1}^{1} z^i \ dz\).

Hint: \(\displaystyle z^i = e^{\log z^i}\).

 

[topsquark]

here is the question

Solve \(\displaystyle \int_{-1}^{1} z^i \ dz\).

Hint: \(\displaystyle z^i = e^{\log z^i}\).

You posted this in "Arithmetic." This is at least Calculus.

So....

What kind of integral is this? Reimann integral? Line integral? Complex integral?

Is i the imaginary unit, or just a constant?

Please make an attempt to do the problem and provide us with the correct context. Then we can start helping you.

-Dan

 

[logistic_guy]

You posted this in "Arithmetic." This is at least Calculus.

So....

What kind of integral is this? Reimann integral? Line integral? Complex integral?

Is i the imaginary unit, or just a constant?

Please make an attempt to do the problem and provide us with the correct context. Then we can start helping you.

-Dan

thank

reiman integral i don't know
what is the difference between line integral and complex integral?

\(\displaystyle i\) same that live in the square root \(\displaystyle i = \sqrt{-1}\)

the answer suppose to be \(\displaystyle \frac{1 + e^{-\pi}}{2}(1 - i)\) i can't get there

i'll show you my attmeb

\(\displaystyle \int_{-1}^{1} z^{i} \ dz = \frac{z^{i + 1}}{i + 1} = \frac{z^{i}z^{1}}{i + 1} = \frac{e^{\log z}z^{1}}{i + 1} = \frac{e^{\log 1}(1)^{1}}{i + 1} - \frac{e^{\log -1}(-1)^{1}}{i + 1}\)