arctan-2-i-i-1-

Question Number 58153 by behi83417@gmail.com last updated on 18/Apr/19

\arctan (\sqrt{2} - i) = ? [i = \sqrt{- 1}]

Answered by MJS last updated on 18/Apr/19

found these :

\tan (a + b i) = \frac{{2 e}^{2 b} \sin 2 a}{{2 e}^{2 b} \cos 2 a + e^{4 b} + 1} + \frac{e^{4 b} - 1}{{2 e}^{2 b} \cos 2 a + e^{4 b} + 1}

\arctan (a + b i) = \frac{1}{2} (π sign a - \arctan \frac{b + 1}{a} + \arctan \frac{b - 1}{a}) + \frac{i}{4} \ln \frac{a^{2} + {(b + 1)}^{2}}{a^{2} + {(b - 1)}^{2}}

\Rightarrow \arctan (\sqrt{2} - i) = \frac{π}{4} + \frac{1}{2} \arctan \frac{\sqrt{2}}{2} - \frac{i}{4} \ln 3 \approx 1.09314 - . 274653 i

Commented by behi83417@gmail.com last updated on 18/Apr/19

d e a r M J S s i r! i t i s a g r e a t k n o w l e d g e!

Commented by MJS last updated on 18/Apr/19

\sin (a + b i) = \sin a \cosh b + i \cos a \sinh b

\cos (a + b i) = \cos a \cosh b - i \sin a \sinh b

\arcsin (a + b i) = \arcsin \frac{\sqrt{a^{2} + b^{2} + 2 a + 1} - \sqrt{a^{2} + b^{2} - 2 a + 1}}{2} + i sign b \ln (\frac{1}{2} (\sqrt{a^{2} + b^{2} - 2 a + 1} + \sqrt{a^{2} + b^{2} + 2 a + 1} + \sqrt{2 (a^{2} + b^{2} - 1 + \sqrt{(a^{2} + b^{2} - 2 a + 1) (a^{2} + b^{2} + 2 a + 1)}}))

\arccos (a + b i) = \arccos \frac{\sqrt{a^{2} + b^{2} + 2 a + 1} - \sqrt{a^{2} + b^{2} - 2 a + 1}}{2} - i sign b \ln (\frac{1}{2} (\sqrt{a^{2} + b^{2} - 2 a + 1} + \sqrt{a^{2} + b^{2} + 2 a + 1} + \sqrt{2 (a^{2} + b^{2} - 1 + \sqrt{(a^{2} + b^{2} - 2 a + 1) (a^{2} + b^{2} + 2 a + 1)}}))

Commented by behi83417@gmail.com last updated on 18/Apr/19

t h a n k s a g a i n s i r .