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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
2 years, 10 months ago

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\(The\quad exponential\quad form\quad of\quad Tan(z)\quad is\\ i\frac { { e }^{ -iz }-{ e }^{ iz } }{ { e }^{ -iz }+{ e }^{ iz } } \\ If\quad z=-\frac { i }{ 2 } Log(\frac { a+ix }{ a-ix } ),\quad then\quad Tan(z)\quad is\\ i({ (\frac { a+ix }{ a-ix } ) }^{ -\frac { 1 }{ 2 } }-{ (\frac { a+ix }{ a-ix } ) }^{ \frac { 1 }{ 2 } })({ (\frac { a+ix }{ a-ix } ) }^{ -\frac { 1 }{ 2 } }+{ (\frac { a+ix }{ a-ix } ) }^{ \frac { 1 }{ 2 } })^{ -1 }\\ Simplifying\quad this\quad reduces\quad it\quad to\quad \frac { x }{ a } \) Michael Mendrin · 2 years, 10 months ago

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@Michael Mendrin Beautiful method!! Shaan Vaidya · 2 years, 10 months ago

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@Michael Mendrin Thanks Michael for the solution I got these result by integrating the function \(\frac{1}{x^2+a^2}\) by two different methods Abdulmuttalib Lokhandwala · 2 years, 10 months ago

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@Abdulmuttalib Lokhandwala As I have mentioned below, it should be \(\frac{a}{x^2+a^2}\) and not \(\frac{1}{x^2+a^2}\) Shaan Vaidya · 2 years, 10 months ago

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@Shaan Vaidya Ya both are true but finally when you do integration a will Already get cancelled and will get the same result Abdulmuttalib Lokhandwala · 2 years, 10 months ago

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@Abdulmuttalib Lokhandwala Vaidya already provided the other method, so I thought I'd include the exponential form route. You know, the brute force way. Michael Mendrin · 2 years, 10 months ago

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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 · 2 years, 10 months ago

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@Shaan Vaidya 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! Calvin Lin Staff · 2 years, 10 months ago

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@Calvin Lin Yes sir I have taken care of the constant c while integrating the function in two ways and they are equal at one point. Abdulmuttalib Lokhandwala · 2 years, 10 months ago

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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, 8 months ago

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@Soham Zemse nice post Abdulmuttalib Lokhandwala · 2 years, 8 months ago

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