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..