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

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

 

[topsquark]

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

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

Please post your questions in an appropriate forum next time.

Actually, it's
[imath]z^i = e^{log(z^i)}[/imath]

So what is [imath]1^i = e^{log(1^i)}[/imath]?

Well, [imath]1^i = r e^{i \theta} \implies r = 1, ~ \theta = 0 \implies 1^i = 1[/imath]

How does this change for [imath](-1)^i[/imath]?

-Dan

 

[logistic_guy]

Please post your questions in an appropriate forum next time.

Actually, it's
[imath]z^i = e^{log(z^i)}[/imath]

So what is [imath]1^i = e^{log(1^i)}[/imath]?

Well, [imath]1^i = r e^{i \theta} \implies r = 1, ~ \theta = 0 \implies 1^i = 1[/imath]

How does this change for [imath](-1)^i[/imath]?

-Dan

\(\displaystyle (-1)^i = re^{i\theta} \to r = 1, \theta = i\pi \to (-1)^i = e^{ii\pi}\)

 

[topsquark]

\(\displaystyle (-1)^i = re^{i\theta} \to r = 1, \theta = i\pi \to (-1)^i = e^{ii\pi}\)

Good. So put the whole thing together and you get...

-Dan

 

[logistic_guy]

Good. So put the whole thing together and you get...

-Dan

thank

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

still not match the answer

 

[topsquark]

thank

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

still not match the answer

[imath]\dfrac{1}{1 + i} ( 1 + e^{-\pi} )[/imath]

Now rationalize the denominator.

-Dan

 

[logistic_guy]

[imath]\dfrac{1}{1 + i} ( 1 + e^{-\pi} )[/imath]

Now rationalize the denominator.

-Dan

i don't understand

 

[The Highlander]

i don't understand

Look at the solution you are trying to reach...

\(\displaystyle \frac{1 + e^{-\pi}}{2}(1 - i)=\color{red}\frac{1 + e^{-\pi}(1-i)}{2}\)​

Now compare that to the result reached so far....

\(\displaystyle \frac{1}{i + 1} + \frac{e^{ii\pi}}{i + 1}=\frac{1}{1 + i} (1 + e^{-\pi})=\color{red}\frac{1+e^{-\pi}}{1 + i}\)​

Considering (only) the numerators of the expressions in red, what would you have to do to the result you have reached (with @topsquark's help) to make it match the desired one?

What else would you need to do if you did that?

(Try it. )

If you don't understand what rationalizing the denominator means then you would do well to look it up (and pay particular attention to conjugates.)

Hope that helps.

 

[logistic_guy]

Look at the solution you are trying to reach...

\(\displaystyle \frac{1 + e^{-\pi}}{2}(1 - i)=\color{red}\frac{1 + e^{-\pi}(1-i)}{2}\)​

Now compare that to the result reached so far....

\(\displaystyle \frac{1}{i + 1} + \frac{e^{ii\pi}}{i + 1}=\frac{1}{1 + i} (1 + e^{-\pi})=\color{red}\frac{1+e^{-\pi}}{1 + i}\)​

Considering (only) the numerators of the expressions in red, what would you have to do to the result you have reached (with @topsquark's help) to make it match the desired one?

What else would you need to do if you did that?

(Try it. )

If you don't understand what rationalizing the denominator means then you would do well to look it up (and pay particular attention to conjugates.)

Hope that helps.

thank The Highlander

why \(\displaystyle \frac{1}{i + 1} + \frac{e^{-\pi}}{i + 1} = \frac{1 + e^{-\pi}}{i + 1}\) and not \(\displaystyle \frac{2 + e^{-\pi}}{i + 1}\)

from where you have \(\displaystyle (1 - i)\)

 

[The Highlander]

thank The Highlander

why \(\displaystyle \frac{1}{i + 1} + \frac{e^{-\pi}}{i + 1} = \frac{1 + e^{-\pi}}{i + 1}\) and not \(\displaystyle \frac{2 + e^{-\pi}}{i + 1}\)

from where you have \(\displaystyle (1 - i)\)

Because: \(\displaystyle \frac{1}{8}+\frac{3}{8}=\frac{1+3}{8}\left(=\frac{4}{8}=\frac{1}{2}\right)\)

If two fractions share the same denominator you simply add their numerators to get the resulting sum!

And: \(\displaystyle 1\text{ plus }e^{-\pi}=1+e^{-\pi}\;not \;2+e^{-\pi}!\)

(I rather expected better of someone operating at your level. )

I "have" \(\displaystyle (1-i)\) straight from the solution you say you are trying to reach in your Post (#3, qv)
(It also happens to be the conjugate of the denominator in the interim result reached so far.)

Hope that helps.

 

[logistic_guy]

Because: \(\displaystyle \frac{1}{8}+\frac{3}{8}=\frac{1+3}{8}\left(=\frac{4}{8}=\frac{1}{2}\right)\)

If two fractions share the same denominator you simply add their numerators to get the resulting sum!

And: \(\displaystyle 1\text{ plus }e^{-\pi}=1+e^{-\pi}\;not \;2+e^{-\pi}!\)

(I rather expected better of someone operating at your level. )

Hope that helps.

my calculation \(\displaystyle \frac{1}{i + 1} + \frac{e^{-\pi}}{i + 1} = \frac{1}{i + 1} + \frac{1}{i + 1}e^{-\pi} = \frac{2}{i + 1}e^{-\pi}\)
because i've a common denomator

(I rather expected better of someone operating at your level. )

is this complement or making fun of me?

I have \(\displaystyle (1-i)\) straight from the solution you say you are trying to reach in your Post (#3, qv)

i know it's from the solution but this is my main issue from where \(\displaystyle (1 - i)\)

 

[The Highlander]

my calculation \(\displaystyle \frac{1}{i + 1} + \frac{e^{-\pi}}{i + 1} = \frac{1}{i + 1} + \frac{1}{i + 1}e^{-\pi} = \frac{2}{i + 1}e^{-\pi}\)
because i've a common denomator

No!!!!
You cannot simply add the two fractions' numerators if you have taken the e^-π out of the second one as a factor!

For example: \(\displaystyle \frac{1}{8}+\frac{15}{8}=\frac{1}{8}+\frac{3\times 5}{8}=\frac{1}{8}+\frac{3}{8}5\color{red}\ne\frac{4}{8}5=\frac{1}{2}5=2½\)​

Because: \(\displaystyle \frac{1}{8}+\frac{15}{8}=\frac{16}{8}=2!!!!\;(Not\;2½)\)


You could say: \(\displaystyle \frac{1}{8}+\frac{15}{8}=\frac{1}{8}+\frac{3\times 5}{8}=\frac{1}{8}+\frac{3}{8}5=0.125+1.875=2\) but you cannot do what you did!

is this complement or making fun of me?

It is certainly not a compliment but neither is it making fun of you! If you are working on a level that involves Calculus then I would expect you to be competent in handling fractions; that is a basic skill that anyone attempting to do Calculus should first possess.

i know it's from the solution but this is my main issue from where \(\displaystyle (1 - i)\)

As I mentioned in my previous posts, as well as appearing in the solution (and, therefore, being what you would like to appear as a factor in the numerator), (1 - i) is also the conjugate of the denominator in the result you had reached aided by @topsquark.
(Rationalizing the denominator is where the (1 - i) really comes from at that stage!)

I suggest that you now go look up "rationalizing the denominator" (Google it?) and you will then understand (I hope) what is being suggested to you.

(You should also brush up on how to add & subtract fractions! )

Hope that helps.

 

[logistic_guy]

No!!!!
You cannot simply add the two fractions' numerators if you have taken the e^-π out of the second one as a factor!

For example: \(\displaystyle \frac{1}{8}+\frac{15}{8}=\frac{1}{8}+\frac{3\times 5}{8}=\frac{1}{8}+\frac{3}{8}5\color{red}\ne\frac{4}{8}5=\frac{1}{2}5=2½\)​

Because: \(\displaystyle \frac{1}{8}+\frac{15}{8}=\frac{16}{8}=2!!!!\;(Not\;2½)\)


You could say: \(\displaystyle \frac{1}{8}+\frac{15}{8}=\frac{1}{8}+\frac{3\times 5}{8}=\frac{1}{8}+\frac{3}{8}5=0.125+1.875=2\) but you cannot do what you did!


It is certainly not a compliment but neither is it making fun of you! If you are working on a level that involves Calculus then I would expect you to be competent in handling fractions; that is a basic skill that anyone attempting to do Calculus should first possess.

As I mentioned in my previous posts, as well as appearing in the solution (and, therefore, being what you would like to appear as a factor in the numerator), (1 - i) is also the conjugate of the denominator in the result you had reached aided by @topsquark.
(Rationalizing the denominator is where the (1 - i) really comes from!)

I suggest that you now go look up "rationalizing the denominator" (Google it?) and you will then understand (I hope) what is being suggested to you.

(You should also brush up on how to add & subtract fractions! )

Hope that helps.

thank

this is too much information. give me sometime to read it

hold on The Highlander, i'll get back to you soon

 

[The Highlander]

thank

this is too much information. give me sometime to read it

hold on The Highlander, i'll get back to you soon

Sorry, it's waaaay past my bedtime!

Just multiply the top and the bottom (of your interim result) by (1 - i) and simplify, then go look up "rationalizing the denominator" (and revise addition & subtraction of fractions) and come back to show us your final algebraic manipulations (to get the answer desired) and confirm that you now understand how that answer was reached.

Good luck!

Hope that helps.

 

[logistic_guy]

No!!!!
You cannot simply add the two fractions' numerators if you have taken the e^-π out of the second one as a factor!

For example: \(\displaystyle \frac{1}{8}+\frac{15}{8}=\frac{1}{8}+\frac{3\times 5}{8}=\frac{1}{8}+\frac{3}{8}5\color{red}\ne\frac{4}{8}5=\frac{1}{2}5=2½\)​

Because: \(\displaystyle \frac{1}{8}+\frac{15}{8}=\frac{16}{8}=2!!!!\;(Not\;2½)\)


You could say: \(\displaystyle \frac{1}{8}+\frac{15}{8}=\frac{1}{8}+\frac{3\times 5}{8}=\frac{1}{8}+\frac{3}{8}5=0.125+1.875=2\) but you cannot do what you did!

i'm back

i think i'm understand this part. thank The Highlander

Rationalize the Denominator - Meaning, Methods, Examples

When we rationalize the denominator in a fraction, then we are eliminating any radical expressions such as square roots and cube roots from the denominator. In this article, let's learn about rationalizing the denominator, its meaning, and methods with solved examples.

www.cuemath.com

i don't understand rationalize. it's using square root

Sorry, it's waaaay past my bedtime!

Just multiply the top and the bottom (of your interim result) by (1 - i) and simplify, then go look up

i'm sorry it's your sleep time

\(\displaystyle \frac{1 + e^{-\pi}}{i + 1}\frac{(1 - i)}{(1 - i)} = \frac{1 - 1i + e^{-\pi}1 - e^{-\pi}i}{i1 - i^2 + 1 - 1i}\)

i need to simply this to get the correct answer, right?

 

[khansaheb]

i'm back
i think i'm understand this part. thank The Highlander

Rationalize the Denominator - Meaning, Methods, Examples

When we rationalize the denominator in a fraction, then we are eliminating any radical expressions such as square roots and cube roots from the denominator. In this article, let's learn about rationalizing the denominator, its meaning, and methods with solved examples.

www.cuemath.com

i don't understand rationalize. it's using square root
i'm sorry it's your sleep time

\(\displaystyle \frac{1 + e^{-\pi}}{i + 1}\frac{(1 - i)}{(1 - i)} = \frac{1 - 1i + e^{-\pi}1 - e^{-\pi}i}{i1 - i^2 + 1 - 1i}\)

i need to simply this to get the correct answer, right?

Almost there

The denominator = (1 + i)*(1 - i) = 1² - (i²) = 1- (-1) = 2

The numerator = (1 + e^-π) * (1 - i) = (1 + e^-π) - i * (1 + e^-π)

In these types of problems, it will be easier to tract if you write the complex expressions as (a + ib) format.

Continue......

 

[The Highlander]

\(\displaystyle \frac{1 + e^{-\pi}}{i + 1}\frac{(1 - i)}{(1 - i)} = \frac{1 - 1i + e^{-\pi}1 - e^{-\pi}i}{i1 - i^2 + 1 - 1i}\)

i need to simply this to get the correct answer, right?

Yes, so why didn't you complete the simplification?

I had hoped that, once you multiplied both top & bottom by (1 - i), you would notice that the numerator became exactly what you were looking for and so it was only the denominator that required expansion & simplification.

\(\displaystyle \frac{1 + e^{-\pi}}{i + 1}\times\frac{(1 - i)}{(1 - i)}\;=\overbrace{\frac{1+e^{-\pi}(1-i)}{(1+i)(1-i)}}^{\text{Just what you want here}}=\;\overbrace{\frac{1+e^{-\pi}(1-i)}{1-i+i-i^2}}^{\text{So leave it alone}}=\frac{1+e^{-\pi}(1-i)}{1-i^2}=\frac{1+e^{-\pi}(1-i)}{1-(ˉ1)}=\frac{1+e^{-\pi}}{2}(1-i)\)​

Notes (on your "simplification"):-

1 is the Multiplicative Identity, ie: when you multiply by 1 whatever you are multiplying stays exactly the same! Therefore, when expanding the likes of (i + 1)(1 - i) you don't need to write: "i1 - i² +1 - 1i", it is preferable (and clearer) to just write: i - i² + 1 - i.

You should also be familiar with the Difference of Squares rule, ie: a² - b² = (a + b)(a - b). So you should have recognized right away that: (1 + i)(1 - i) = 1 - i² and there is, therefore, no need to expand those brackets into 4 separate terms thereby creating an extra (unnecessary) step!

Do you also now recognize the reason for multiplying by (1 - i)? Rationalizing the denominator removes any square roots (radicals) that are in the denominator as it is undesirable to have radicals there. i is, of course, a square root (\(\displaystyle \sqrt{-1}\)) and, therefore, to eliminate it, you multiply by the conjugate of (1 + i) which, as you now know (), is (1 - i) because (1 + i)(1 - i) expands to 1 - i² = 1 - \(\displaystyle \sqrt{-1}^2\) = 1 - (ˉ1) = 2.

Hope that helps.

 

[logistic_guy]

Almost there

The denominator = (1 + i)*(1 - i) = 1² - (i²) = 1- (-1) = 2

The numerator = (1 + e^-π) * (1 - i) = (1 + e^-π) - i * (1 + e^-π)

In these types of problems, it will be easier to tract if you write the complex expressions as (a + ib) format.

Continue......

thank

Yes, so why didn't you complete the simplification?

i do but i was stuck in the numator

\(\displaystyle \frac{1 + e^{-\pi}}{i + 1}\times\frac{(1 - i)}{(1 - i)}\;=\overbrace{\frac{1+e^{-\pi}(1-i)}{(1+i)(1-i)}}^{\text{Just what you want here}}=\;\overbrace{\frac{1+e^{-\pi}(1-i)}{1-i+i-i^2}}^{\text{So leave it alone}}=\frac{1+e^{-\pi}(1-i)}{1-i^2}=\frac{1+e^{-\pi}(1-i)}{1-(ˉ1)}=\frac{1+e^{-\pi}}{2}(1-i)\)​

i think i'm understand. i was stupid to multiply the factors of numator \(\displaystyle (1 + e^{-\pi})(i - 1)\)
i shouldn't multiply them

Notes (on your "simplification"):-

1 is the Multiplicative Identity, ie: when you multiply by 1 whatever you are multiplying stays exactly the same! Therefore, when expanding the likes of (i + 1)(1 - i) you don't need to write: "i1 - i² +1 - 1i", it is preferable (and clearer) to just write: i - i² + 1 - i.

i do that because i want to show you my multiplication

You should also be familiar with the Difference of Squares rule, ie: a² - b² = (a + b)(a - b). So you should have recognized right away that: (1 + i)(1 - i) = 1 - i² and there is, therefore, no need to expand those brackets into 4 separate terms thereby creating an extra (unnecessary) step!

this i learn in algebra and i think i'm very good in algebra now. i'm understand

Do you also now recognize the reason for multiplying by (1 - i)? Rationalizing the denominator removes any square roots (radicals) that are in the denominator as it is undesirable to have radicals there. i is, of course, a square root (\(\displaystyle \sqrt{-1}\)) and, therefore, to eliminate it, you multiply by the conjugate of (1 + i) which, as you now know (), is (1 - i) because (1 + i)(1 - i) expands to 1 - i² = 1 - \(\displaystyle \sqrt{-1}^2\) = 1 - (ˉ1) = 2.

i know before \(\displaystyle i = \sqrt{-1}\) but i'm understanding now it's related to conjugate

the link i give in my post 17 say the conjugate of \(\displaystyle 7 + \sqrt{5}\) is \(\displaystyle 7 - \sqrt{5}\) but we do the conjugate of \(\displaystyle i + 1\) is \(\displaystyle 1 - i\) not \(\displaystyle i - 1\) why?

see my calulcation
\(\displaystyle \frac{(1 + e^{-\pi})(1 - i)}{(i + 1)(1 - i)} = \frac{(1 + e^{-\pi})(1 - i)}{i - i^2 + 1 - i} = \frac{(1 + e^{-\pi})(1 - i)}{-i^2 + 1} = \frac{(1 + e^{-\pi})(1 - i)}{-(-1) + 1} = \frac{(1 + e^{-\pi})(1 - i)}{+1 + 1} = \frac{(1 + e^{-\pi})(1 - i)}{2} = \frac{1 + e^{-\pi}}{2}(1 - i)\) it match the answer i want

Hope that helps.

yes it help very much and i learn too much form your nice explanation. thank very much The Highlander