solve-for-z-in-the-form-x-iy-if-tanz-0-5-

Question Number 50849 by peter frank last updated on 21/Dec/18

solve f o r z i n t h e f o r m x + i y

i f t a n z = 0.5

Answered by ajfour last updated on 21/Dec/18

e^{i z} = \cos z + i \sin z

e^{- i z} = \cos z - i \sin z

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

l e t e^{i z} = e^{- y + i x} = e^{- y} (\cos x + i \sin x)

\Rightarrow e^{- i z} = e^{y - i x} = e^{y} (\cos x - i \sin x)

\Rightarrow \tan z = \frac{(e^{- y} - e^{y}) \cos x + i (e^{- y} + e^{y}) \sin x}{i [(e^{- y} + e^{y}) \cos x + i (e^{- y} - e^{y}) \sin x]}

l e t e^{- y} + e^{y} = u, e^{- y} - e^{y} = v

\tan z = \frac{(- i v \cos x + u \sin x) (u \cos x - i v \sin x)}{u^{2} \cos^{2} x + v^{2} \sin^{2} x}

\tan z = \frac{- i u v + (u^{2} - v^{2}) \sin 2 x}{2 (u^{2} \cos^{2} x + v^{2} \sin^{2} x)}

i f \tan z = \frac{1}{2}, \Rightarrow

u v = 0 o r e^{- 2 y} = e^{2 y}

\Rightarrow y = 0

\Rightarrow u = 2, v = 0

\frac{(u^{2} - v^{2}) \sin 2 x}{2 (u^{2} \cos^{2} x + v^{2} \sin^{2} x)} = \frac{1}{2}

\Rightarrow \tan x = \frac{1}{2}

\Rightarrow z = x = \tan^{- 1} \frac{1}{2} + n π .

Commented by peter frank last updated on 21/Dec/18

a n s [z = \frac{π n}{2} + 26.56 + 0.00000119 i]

Commented by MJS last updated on 21/Dec/18

sorry but this is wrong . possibly the calculator

approximates in a crazy way .

\tan (a + b i) = \frac{1}{2}

has only real solutions \Rightarrow b = 0

Commented by peter frank last updated on 22/Dec/18

t h a n k y o u b o t h s i r s

Answered by MJS last updated on 21/Dec/18

\tan (x + i y) = \frac{1}{2}; (\begin{matrix} x \\ y \end{matrix}) \in R^{2}

x + i y = n π + \arctan \frac{1}{2}; n \in Z

\Rightarrow y = 0; x = n π + \arctan \frac{1}{2}; n \in Z

Answered by peter frank last updated on 21/Dec/18

\tanh (i z) = i t a n z

\frac{1}{2} = \frac{\tanh (i z)}{i}

\frac{1}{2} = \frac{e^{i z} - e^{- i z}}{i (e^{i z} + e^{- i z})}

m u l t i p l y b y e^{i z} a n d s i m p l i f y

e^{2 i z} = \frac{3}{5} + \frac{4 i}{5}

z = x + i y

e^{2 i x} . e^{- 2 y} = \frac{3}{5} + \frac{4 i}{5}

f r o m {E u r e l}^{'} s f o r m o f

c o m p l e x n o

e^{2 x i} = \cos 2 x + i \sin 2 x

e^{- 2 y} (\cos 2 x + i \sin 2 x) = \frac{3}{4} + \frac{4 i}{5}

c o m p a r e r e a l a n d I m m a r g i n a r y

e^{- 2 y} \cos 2 x = \frac{3}{5} \dots . (i)

e^{- 2 y} \sin 2 x = \frac{4}{5} \dots . (i i)

d i v i d e i i b y i

t a n 2 x = \frac{4}{3}

t a n 2 x = \tan (53.13010235) \approx \tan 53.13

2 x = π n + 53 . 13^{°}

x = \frac{π n}{2} + 26.565

n = 0, 1, 2, 3

\dots .

f r o m

e^{- 2 y} \cos 2 x = \frac{3}{5} \dots . (i)

e^{- 2 y} = \frac{0.6}{\cos 2 x}

- 2 y = l n (\frac{0.6}{\cos 2 x})

y = \frac{1}{2} l n (\frac{0.6}{\cos 2 x})

z = x + i y

z = \frac{π n}{2} + 26.565 - \frac{1}{2} l n (\frac{0.6}{\cos 2 x})

\dots \dots