# complex trigonometry??

Prove that -$\frac{i}{2}$Ln($\frac{a+ix}{a-ix}$) = arctan($\frac{x}{a}$)

i = $\sqrt{-1}$

Post the solution if you have solved it

Note by Abdulmuttalib Lokhandwala
7 years, 2 months ago

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Differentiate both the sides individually. Doing that shall yield $\frac{a}{a^{2}+x^{2}}$ on both sides,

- 7 years, 2 months ago

Note that you still have to show that they are equal at one point, otherwise the graphs could be vertical shifts of each other.

Don't forget your constant $+ C$ when integrating!

Staff - 7 years, 2 months ago

Yes sir I have taken care of the constant c while integrating the function in two ways and they are equal at one point.

- 7 years, 2 months ago

$The\quad exponential\quad form\quad of\quad Tan(z)\quad is\\ i\frac { { e }^{ -iz }-{ e }^{ iz } }{ { e }^{ -iz }+{ e }^{ iz } } \\ If\quad z=-\frac { i }{ 2 } Log(\frac { a+ix }{ a-ix } ),\quad then\quad Tan(z)\quad is\\ i({ (\frac { a+ix }{ a-ix } ) }^{ -\frac { 1 }{ 2 } }-{ (\frac { a+ix }{ a-ix } ) }^{ \frac { 1 }{ 2 } })({ (\frac { a+ix }{ a-ix } ) }^{ -\frac { 1 }{ 2 } }+{ (\frac { a+ix }{ a-ix } ) }^{ \frac { 1 }{ 2 } })^{ -1 }\\ Simplifying\quad this\quad reduces\quad it\quad to\quad \frac { x }{ a }$

- 7 years, 2 months ago

Thanks Michael for the solution I got these result by integrating the function $\frac{1}{x^2+a^2}$ by two different methods

- 7 years, 2 months ago

Vaidya already provided the other method, so I thought I'd include the exponential form route. You know, the brute force way.

- 7 years, 2 months ago

As I have mentioned below, it should be $\frac{a}{x^2+a^2}$ and not $\frac{1}{x^2+a^2}$

- 7 years, 2 months ago

Ya both are true but finally when you do integration a will Already get cancelled and will get the same result

- 7 years, 2 months ago

Beautiful method!!

- 7 years, 2 months ago

- 7 years, 1 month ago

nice post

- 7 years, 1 month ago