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

i = \(\sqrt{-1}\)

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

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Beautiful method!!

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

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

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

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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.

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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!

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

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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

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nice post

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