solve-for-x-C-cos-x-a-bi-

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

\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}) \dots n o w i t s

e a z y w i t h t h a t \dots