arctan((√2)−i)=? [i=(√(−1))]


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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

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