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

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

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TopNewest\(The\quad exponential\quad form\quad of\quad Tan(z)\quad is\\ i\frac { { e }^{ -iz }-{ e }^{ iz } }{ { e }^{ -iz }+{ e }^{ iz } } \\ If\quad z=-\frac { i }{ 2 } Log(\frac { a+ix }{ a-ix } ),\quad then\quad Tan(z)\quad is\\ i({ (\frac { a+ix }{ a-ix } ) }^{ -\frac { 1 }{ 2 } }-{ (\frac { a+ix }{ a-ix } ) }^{ \frac { 1 }{ 2 } })({ (\frac { a+ix }{ a-ix } ) }^{ -\frac { 1 }{ 2 } }+{ (\frac { a+ix }{ a-ix } ) }^{ \frac { 1 }{ 2 } })^{ -1 }\\ Simplifying\quad this\quad reduces\quad it\quad to\quad \frac { x }{ a } \) – Michael Mendrin · 3 years ago

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– Shaan Vaidya · 3 years ago

Beautiful method!!Log in to reply

– Abdulmuttalib Lokhandwala · 3 years ago

Thanks Michael for the solution I got these result by integrating the function \(\frac{1}{x^2+a^2}\) by two different methodsLog in to reply

– Shaan Vaidya · 3 years ago

As I have mentioned below, it should be \(\frac{a}{x^2+a^2}\) and not \(\frac{1}{x^2+a^2}\)Log in to reply

– Abdulmuttalib Lokhandwala · 3 years ago

Ya both are true but finally when you do integration a will Already get cancelled and will get the same resultLog in to reply

– Michael Mendrin · 3 years ago

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

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

Thus, your conjecture is proved. – Shaan Vaidya · 3 years ago

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Don't forget your constant \( + C \) when integrating! – Calvin Lin Staff · 3 years ago

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– Abdulmuttalib Lokhandwala · 3 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.Log in to reply

please do read my post at : https://brilliant.org/discussions/thread/math-is-getting-broken/ it is related to this question. thanks – Soham Zemse · 2 years, 11 months ago

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– Abdulmuttalib Lokhandwala · 2 years, 11 months ago

nice postLog in to reply