# 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
4 years ago

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

- 4 years ago

Beautiful method!!

- 4 years ago

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

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

- 4 years ago

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

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

- 4 years ago

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

- 4 years 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 - 4 years 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.

- 3 years, 10 months ago

nice post

- 3 years, 10 months ago