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

what are the complex solution tan  (z) = −2 ?

Commented bymr W last updated on 15/Jul/20

z=kπ−tan^(−1) 2  z is always real!

Commented bybemath last updated on 15/Jul/20

nothing in complex sir?

Commented bymr W last updated on 15/Jul/20

a real number is also a complex  number. if you want,  z=kπ−tan^(−1) 2+0i

Commented bybobhans last updated on 15/Jul/20

z = kπ−tan^(−1) (2) or z=kπ−tan^(−1) (−2)  sir?

Commented bymr W last updated on 15/Jul/20

z = kπ−tan^(−1) (2)

Commented bybramlex last updated on 15/Jul/20

tan (z) = −2 ⇒ tan (−z)=2  note tan x = tan θ ⇒x = θ+nπ  ⇒ tan (−z)= tan (tan^(−1) (2))  −z+nπ = tan^(−1) (2)   z = nπ−tan^(−1) (2) (⊕)

Answered by OlafThorendsen last updated on 15/Jul/20

tanz = −2  ((sinz)/(cosz)) = −2  (((e^(iz) −e^(−iz) )/(2i))/((e^(iz) +e^(−iz) )/2)) = −2  ((e^(iz) −e^(−iz) )/(e^(iz) +e^(−iz) )) = −2i  e^(iz) −e^(−iz)  = −2i(e^(iz) +e^(−iz) )  (1+2i)e^(iz)  = (1−2i)e^(−iz)   (1+2i)e^(2iz)  = (1−2i)  e^(2iz)  = ((1−2i)/(1+2i)) = (((1−2i)^2 )/5) = −((3+4i)/5)  e^(2iz)  = e^(i(Arctan(4/3)+2kπ))   z = (1/2)Arctan(4/3)+kπ, k∈Z  Note :  ((1/2)Arctan(4/3)−π = Arctan(−2))

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

tanz =−2 ⇒((sinz)/(cosz))=−2 ⇒sinz =−2cosz ⇒  sin^2 z =4cos^2 z ⇒1−cos^2 z =4cos^2 z ⇒1 =5cos^2 z ⇒  cos^2 z =(1/5) ⇒cosz =+^− (1/(√5))  cosz =(1/(√5)) ⇒ch(iz) =(1/(√5)) ⇒ ((e^(iz)  +e^(−iz) )/2) =(1/(√5)) ⇒e^(iz)  +e^(−iz)  =(2/(√5))  e^(iz)  =t ⇒t+t^(−1)  =(2/(√5)) ⇒t^2  +1 =(2/(√5))t ⇒t^2  −((2t)/(√5)) +1 =0 ⇒  (√5)t^2 −2t +(√5)=0 →Δ^′  =1−5 =−4 ⇒t_1 =1+2i and t_2 =1−2i   e^(iz)  =t ⇒iz =lnt ⇒z =−iln(t) ⇒z =−iln(1+^− 2i)  case 2  cosz =−(1/(√5)) ⇒((e^(iz)  +e^(−iz) )/2) =−(1/(√5)) ⇒  e^(iz)  +e^(−iz)   =((−2)/(√5)) ⇒t +t^(−1 )  =−(2/(√5))  (t =e^(iz) ) ⇒  t^2  +1 =−(2/(√5))t ⇒t^2 +((2t)/(√5)) +1 =0 ⇒(√5)t^2 +2t +(√5)=0 →Δ^′  =1−5 =−4 ⇒  t_1 =−1+2i and t_2 =−1−2i  z =−iln(t) ⇒z =−iln(−1 +^− 2i)

Commented bymr W last updated on 15/Jul/20

but t_1 =−1+2i=e^((π−tan^(−1) 2)i)   z=−iln t_1 =ln t_1 ^(−i) =ln e^((π−tan^(−1) 2)i(−i))   =π−tan^(−1) 2

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

thank you sir i havent verify..