what-are-the-complex-solution-tan-z-2-

Question Number 103459 by bemath last updated on 15/Jul/20

w h a t a r e t h e c o m p l e x s o l u t i o n \tan

(z) = - 2 ?

Commented by mr W last updated on 15/Jul/20

z = k π - \tan^{- 1} 2

z i s a l w a y s r e a l!

Commented by bemath last updated on 15/Jul/20

n o t h i n g i n c o m p l e x s i r ?

Commented by mr W last updated on 15/Jul/20

a r e a l n u m b e r i s a l s o a c o m p l e x

n u m b e r . i f y o u w a n t,

z = k π - \tan^{- 1} 2 + 0 i

Commented by bobhans last updated on 15/Jul/20

z = k π - \tan^{- 1} (2) o r z = k π - \tan^{- 1} (- 2)

s i r ?

Commented by mr W last updated on 15/Jul/20

z = k π - \tan^{- 1} (2)

Commented by bramlex last updated on 15/Jul/20

\tan (z) = - 2 \Rightarrow \tan (- z) = 2

note \tan x = \tan θ \Rightarrow x = θ + n π

\Rightarrow \tan (- z) = \tan (\tan^{- 1} (2))

- z + n π = \tan^{- 1} (2)

z = n π - \tan^{- 1} (2) (\oplus)

Answered by OlafThorendsen last updated on 15/Jul/20

\tan z = - 2

\frac{\sin z}{\cos z} = - 2

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

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

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

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

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

e^{2 i z} = \frac{1 - 2 i}{1 + 2 i} = \frac{{(1 - 2 i)}^{2}}{5} = - \frac{3 + 4 i}{5}

e^{2 i z} = e^{i (Arctan \frac{4}{3} + 2 k π)}

z = \frac{1}{2} Arctan \frac{4}{3} + k π, k \in Z

Note :

(\frac{1}{2} Arctan \frac{4}{3} - π = Arctan (- 2))

Answered by mathmax by abdo last updated on 15/Jul/20

tanz = - 2 \Rightarrow \frac{sinz}{cosz} = - 2 \Rightarrow sinz = - 2 cosz \Rightarrow

\sin^{2} z = {4 \cos}^{2} z \Rightarrow 1 - \cos^{2} z = {4 \cos}^{2} z \Rightarrow 1 = {5 \cos}^{2} z \Rightarrow

\cos^{2} z = \frac{1}{5} \Rightarrow cosz = \bar{+} \frac{1}{\sqrt{5}}

cosz = \frac{1}{\sqrt{5}} \Rightarrow ch (iz) = \frac{1}{\sqrt{5}} \Rightarrow \frac{e^{iz} + e^{- iz}}{2} = \frac{1}{\sqrt{5}} \Rightarrow e^{iz} + e^{- iz} = \frac{2}{\sqrt{5}}

e^{iz} = t \Rightarrow t + t^{- 1} = \frac{2}{\sqrt{5}} \Rightarrow t^{2} + 1 = \frac{2}{\sqrt{5}} t \Rightarrow t^{2} - \frac{2 t}{\sqrt{5}} + 1 = 0 \Rightarrow

\sqrt{5} t^{2} - 2 t + \sqrt{5} = 0 \to Δ^{^{'}} = 1 - 5 = - 4 \Rightarrow t_{1} = 1 + 2 i and t_{2} = 1 - 2 i

e^{iz} = t \Rightarrow iz = lnt \Rightarrow z = - iln (t) \Rightarrow z = - iln (1 \bar{+} 2 i)

case 2 cosz = - \frac{1}{\sqrt{5}} \Rightarrow \frac{e^{iz} + e^{- iz}}{2} = - \frac{1}{\sqrt{5}} \Rightarrow

e^{iz} + e^{- iz} = \frac{- 2}{\sqrt{5}} \Rightarrow t + t^{- 1} = - \frac{2}{\sqrt{5}} (t = e^{iz}) \Rightarrow

t^{2} + 1 = - \frac{2}{\sqrt{5}} t \Rightarrow t^{2} + \frac{2 t}{\sqrt{5}} + 1 = 0 \Rightarrow \sqrt{5} t^{2} + 2 t + \sqrt{5} = 0 \to Δ^{^{'}} = 1 - 5 = - 4 \Rightarrow

t_{1} = - 1 + 2 i and t_{2} = - 1 - 2 i

z = - iln (t) \Rightarrow z = - iln (- 1 \bar{+} 2 i)

Commented by mr W last updated on 15/Jul/20

b u t t_{1} = - 1 + 2 i = e^{(π - \tan^{- 1} 2) i}

z = - i \ln t_{1} = \ln t_{1}^{- i} = \ln e^{(π - \tan^{- 1} 2) i (- i)}

= π - \tan^{- 1} 2

Commented by mathmax by abdo last updated on 16/Jul/20

thank you sir i havent verify . .