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Question Number 116560 by Bird last updated on 04/Oct/20

i f a r c t a n (x + i y) = a + i b

w i t h a a n d b r e a l s d e t e r m i n e

a a n d b

Commented by MJS_new last updated on 05/Oct/20

\arctan (x + i y) = a + i b

a = \frac{π}{2} sign x + \frac{1}{2} (\arctan \frac{y - 1}{x} - \arctan \frac{y + 1}{x})

b = \frac{1}{4} \ln \frac{x^{2} + {(y + 1)}^{2}}{x^{2} + {(y - 1)}^{2}}

if x = 0

a = real (\frac{i}{2} \ln \frac{1 + y}{1 - y})

b = imag (\frac{i}{2} \ln \frac{1 + y}{1 - y})

Answered by Olaf last updated on 05/Oct/20

x + i y = \tan (a + i b)

x + i y = \frac{\tan a + \tan i b}{1 - \tan a \tan i b}

\tan i b = \frac{\sin i b}{\cos i b}

\tan i b = \frac{\frac{e^{i (i b)} - e^{- i (i b)}}{2 i}}{\frac{e^{i (i b)} + e^{- i (i b)}}{2}} = - i \frac{e^{- b} - e^{i b}}{e^{- i b} + e^{i b}}

\tan i b = i \frac{e^{b} - e^{- i b}}{e^{i b} + e^{- i b}} = i \frac{\sinh b}{\cosh b} = i \tanh b

x + i y = \frac{\tan a + i \tanh b}{1 - i \tan a \tanh b}

x + i y = \frac{(\tan a + i \tanh b) (1 + i \tan a \tanh b)}{1 + \tan^{2} a \tanh^{2} b}

x + i y = \frac{\tan a (1 - \tanh^{2} b) + i \tanh b (1 + \tan^{2} a)}{1 + \tan^{2} a \tanh^{2} b}

\frac{y}{x} = \frac{\tanh b (1 + \tan^{2} a)}{\tan a (1 - \tanh^{2} b)}

\frac{y}{x} = \frac{\tanh b \cosh^{2} b}{\tan a \cos^{2} a} = \frac{\sinh b \cosh b}{\sin a \cos a}

\frac{y}{x} = \frac{\tanh b \cosh^{2} b}{\tan a \cos^{2} a} = \frac{\sinh 2 b}{\sin 2 a} (1)

x^{2} + y^{2} = \frac{\tan^{2} a}{\cosh^{4} b} + \frac{\tanh^{2} b}{\cos^{4} a}

x^{2} + y^{2} = \frac{\sin^{2} a \cos^{2} a + \sinh^{2} b \cosh^{2} b}{\cos^{4} a \cosh^{4} b}

x^{2} + y^{2} = \frac{\sin^{2} 2 a + \sinh^{2} 2 b}{{4 \cos}^{4} a \cosh^{4} b} (2)

work in progress \dots

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