solve for x∈C cos (x)=a+bi


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Question Number 88594 by mr W last updated on 11/Apr/20

$s o l v e f o r x \in C$ $\cos (x) = a + b i$
Commented by abdomathmax last updated on 11/Apr/20
$⇒ch(ix) =a+bi ⇒((e^(ix) +e^(−ix) )/2)=a+bi ⇒ e^(ix) +e^(−ix) =2a+2bi let e^(ix) =z ⇒ix =ln(z) ⇒ x =−iln(z) (e)→z+z^(−1) =2a+2bi ⇒z^2 +1 =(2a+2bi)z ⇒ z^2 −2(a+bi)z +1 =0 Δ^′ =(a+bi)^2 −1 =a^2 +2abi−b^2 −1 =a^2 −b^2 −1 +2abi =(√((a^2 −b^2 −1)^2 +4a^2 b^2 ))e^(iarctan(((2ab)/(a^2 −b^2 −1)))) ⇒z_1 =((a+bi +((a^2 −b^2 −1)^2 +4a^2 b^2 )^(1/4) e^((i/2)arcran(((2ab)/(a^2 −b^2 −1)))) )/1) z_2 =a+bi+{(a^2 −b^2 −1)^2 +4a^2 b^2 }^(1/4) e^((i/2) arctan(((2ab)/(a^2 −b^2 −1)))) ⇒x =−iln( a+bi +^− {(a^2 −b^2 −1)^2 +4a^2 b^2 }^(1/4) e^((i/2)arctan(((2ab)/(a^2 −b^2 −1))))$
$\Rightarrow c h (i x) = a + b i \Rightarrow \frac{e^{i x} + e^{- i x}}{2} = a + b i \Rightarrow$ $e^{i x} + e^{- i x} = 2 a + 2 b i l e t e^{i x} = z \Rightarrow i x = l n (z) \Rightarrow$ $x = - i l n (z)$ $(e) \to z + z^{- 1} = 2 a + 2 b i \Rightarrow z^{2} + 1 = (2 a + 2 b i) z \Rightarrow$ $z^{2} - 2 (a + b i) z + 1 = 0$ $Δ^{^{'}} = {(a + b i)}^{2} - 1 = a^{2} + 2 a b i - b^{2} - 1$ $= a^{2} - b^{2} - 1 + 2 a b i = \sqrt{{(a^{2} - b^{2} - 1)}^{2} + 4 a^{2} b^{2}} e^{i a r c t a n (\frac{2 a b}{a^{2} - b^{2} - 1})}$ $\Rightarrow z_{1} = \frac{a + b i + {({(a^{2} - b^{2} - 1)}^{2} + 4 a^{2} b^{2})}^{\frac{1}{4}} e^{\frac{i}{2} a r c r a n (\frac{2 a b}{a^{2} - b^{2} - 1})}}{1}$ $z_{2} = a + b i + {{(a^{2} - b^{2} - 1)}^{2} + 4 a^{2} b^{2}}^{\frac{1}{4}} e^{\frac{i}{2} a r c t a n (\frac{2 a b}{a^{2} - b^{2} - 1})}$ $\Rightarrow x = - i l n (a + b i \bar{+} {{(a^{2} - b^{2} - 1)}^{2} + 4 a^{2} b^{2}}^{\frac{1}{4}} e^{\frac{i}{2} a r c t a n (\frac{2 a b}{a^{2} - b^{2} - 1})}$
Commented by mr W last updated on 11/Apr/20

$t h a n k y o u s i r! c a n y o u p l e a s e p u t i t i n t o$ $x + y i f o r m ?$
Commented by abdomathmax last updated on 11/Apr/20

$y o u a r e w e l c o m e y o u c a n s i r u s e$ $u + i v = \sqrt{u^{2} + v^{2}} e^{i s r c t a n (\frac{v}{u})} \Rightarrow$ $l n (u + i v) = \frac{1}{2} l n (u^{2} + v^{2}) + i a r c t a n (\frac{v}{u}) . . . n o w i t s$ $e a z y w i t h t h a t . . .$
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