Menu Close

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

Tinku Tara 0 Comments
Pin




Question Number 103459 by bemath last updated on 15/Jul/20
what are the complex solution tan (z) = −2 ?
$${what}\:{are}\:{the}\:{complex}\:{solution}\:\mathrm{tan} \\ $$$$\left({z}\right)\:=\:−\mathrm{2}\:? \\ $$
Commented by mr W last updated on 15/Jul/20
z=kπ−tan^(−1) 2 z is always real!
$${z}={k}\pi−\mathrm{tan}^{−\mathrm{1}} \mathrm{2} \\ $$$${z}\:{is}\:{always}\:{real}! \\ $$
Commented by bemath last updated on 15/Jul/20
nothing in complex sir?
$${nothing}\:{in}\:{complex}\:{sir}? \\ $$
Commented by mr W last updated on 15/Jul/20
a real number is also a complex number. if you want, z=kπ−tan^(−1) 2+0i
$${a}\:{real}\:{number}\:{is}\:{also}\:{a}\:{complex} \\ $$$${number}.\:{if}\:{you}\:{want}, \\ $$$${z}={k}\pi−\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+\mathrm{0}{i} \\ $$
Commented by bobhans last updated on 15/Jul/20
z = kπ−tan^(−1) (2) or z=kπ−tan^(−1) (−2) sir?
$${z}\:=\:{k}\pi−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)\:{or}\:{z}={k}\pi−\mathrm{tan}^{−\mathrm{1}} \left(−\mathrm{2}\right) \\ $$$${sir}? \\ $$
Commented by mr W last updated on 15/Jul/20
z = kπ−tan^(−1) (2)
$${z}\:=\:{k}\pi−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right) \\ $$
Commented by bramlex 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) (⊕)
$$\mathrm{tan}\:\left(\mathrm{z}\right)\:=\:−\mathrm{2}\:\Rightarrow\:\mathrm{tan}\:\left(−\mathrm{z}\right)=\mathrm{2} \\ $$$$\mathrm{note}\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{tan}\:\theta\:\Rightarrow\mathrm{x}\:=\:\theta+\mathrm{n}\pi \\ $$$$\Rightarrow\:\mathrm{tan}\:\left(−\mathrm{z}\right)=\:\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)\right) \\ $$$$−\mathrm{z}+\mathrm{n}\pi\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)\: \\ $$$$\mathrm{z}\:=\:\mathrm{n}\pi−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)\:\left(\oplus\right) \\ $$
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))
$$\mathrm{tan}{z}\:=\:−\mathrm{2} \\ $$$$\frac{\mathrm{sin}{z}}{\mathrm{cos}{z}}\:=\:−\mathrm{2} \\ $$$$\frac{\frac{{e}^{{iz}} −{e}^{−{iz}} }{\mathrm{2}{i}}}{\frac{{e}^{{iz}} +{e}^{−{iz}} }{\mathrm{2}}}\:=\:−\mathrm{2} \\ $$$$\frac{{e}^{{iz}} −{e}^{−{iz}} }{{e}^{{iz}} +{e}^{−{iz}} }\:=\:−\mathrm{2}{i} \\ $$$${e}^{{iz}} −{e}^{−{iz}} \:=\:−\mathrm{2}{i}\left({e}^{{iz}} +{e}^{−{iz}} \right) \\ $$$$\left(\mathrm{1}+\mathrm{2}{i}\right){e}^{{iz}} \:=\:\left(\mathrm{1}−\mathrm{2}{i}\right){e}^{−{iz}} \\ $$$$\left(\mathrm{1}+\mathrm{2}{i}\right){e}^{\mathrm{2}{iz}} \:=\:\left(\mathrm{1}−\mathrm{2}{i}\right) \\ $$$${e}^{\mathrm{2}{iz}} \:=\:\frac{\mathrm{1}−\mathrm{2}{i}}{\mathrm{1}+\mathrm{2}{i}}\:=\:\frac{\left(\mathrm{1}−\mathrm{2}{i}\right)^{\mathrm{2}} }{\mathrm{5}}\:=\:−\frac{\mathrm{3}+\mathrm{4}{i}}{\mathrm{5}} \\ $$$${e}^{\mathrm{2}{iz}} \:=\:{e}^{{i}\left(\mathrm{Arctan}\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{2}{k}\pi\right)} \\ $$$${z}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Arctan}\frac{\mathrm{4}}{\mathrm{3}}+{k}\pi,\:{k}\in\mathbb{Z} \\ $$$$\mathrm{Note}\:: \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Arctan}\frac{\mathrm{4}}{\mathrm{3}}−\pi\:=\:\mathrm{Arctan}\left(−\mathrm{2}\right)\right) \\ $$
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)
$$\mathrm{tanz}\:=−\mathrm{2}\:\Rightarrow\frac{\mathrm{sinz}}{\mathrm{cosz}}=−\mathrm{2}\:\Rightarrow\mathrm{sinz}\:=−\mathrm{2cosz}\:\Rightarrow \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{z}\:=\mathrm{4cos}^{\mathrm{2}} \mathrm{z}\:\Rightarrow\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{z}\:=\mathrm{4cos}^{\mathrm{2}} \mathrm{z}\:\Rightarrow\mathrm{1}\:=\mathrm{5cos}^{\mathrm{2}} \mathrm{z}\:\Rightarrow \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{z}\:=\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow\mathrm{cosz}\:=\overset{−} {+}\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{cosz}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{ch}\left(\mathrm{iz}\right)\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\:\frac{\mathrm{e}^{\mathrm{iz}} \:+\mathrm{e}^{−\mathrm{iz}} }{\mathrm{2}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{e}^{\mathrm{iz}} \:+\mathrm{e}^{−\mathrm{iz}} \:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{e}^{\mathrm{iz}} \:=\mathrm{t}\:\Rightarrow\mathrm{t}+\mathrm{t}^{−\mathrm{1}} \:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\mathrm{t}\:\Rightarrow\mathrm{t}^{\mathrm{2}} \:−\frac{\mathrm{2t}}{\:\sqrt{\mathrm{5}}}\:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow \\ $$$$\sqrt{\mathrm{5}}\mathrm{t}^{\mathrm{2}} −\mathrm{2t}\:+\sqrt{\mathrm{5}}=\mathrm{0}\:\rightarrow\Delta^{'} \:=\mathrm{1}−\mathrm{5}\:=−\mathrm{4}\:\Rightarrow\mathrm{t}_{\mathrm{1}} =\mathrm{1}+\mathrm{2i}\:\mathrm{and}\:\mathrm{t}_{\mathrm{2}} =\mathrm{1}−\mathrm{2i}\: \\ $$$$\mathrm{e}^{\mathrm{iz}} \:=\mathrm{t}\:\Rightarrow\mathrm{iz}\:=\mathrm{lnt}\:\Rightarrow\mathrm{z}\:=−\mathrm{iln}\left(\mathrm{t}\right)\:\Rightarrow\mathrm{z}\:=−\mathrm{iln}\left(\mathrm{1}\overset{−} {+}\mathrm{2i}\right) \\ $$$$\mathrm{case}\:\mathrm{2}\:\:\mathrm{cosz}\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\frac{\mathrm{e}^{\mathrm{iz}} \:+\mathrm{e}^{−\mathrm{iz}} }{\mathrm{2}}\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow \\ $$$$\mathrm{e}^{\mathrm{iz}} \:+\mathrm{e}^{−\mathrm{iz}} \:\:=\frac{−\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{t}\:+\mathrm{t}^{−\mathrm{1}\:} \:=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\:\left(\mathrm{t}\:=\mathrm{e}^{\mathrm{iz}} \right)\:\Rightarrow \\ $$$$\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\:=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\mathrm{t}\:\Rightarrow\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{2t}}{\:\sqrt{\mathrm{5}}}\:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\sqrt{\mathrm{5}}\mathrm{t}^{\mathrm{2}} +\mathrm{2t}\:+\sqrt{\mathrm{5}}=\mathrm{0}\:\rightarrow\Delta^{'} \:=\mathrm{1}−\mathrm{5}\:=−\mathrm{4}\:\Rightarrow \\ $$$$\mathrm{t}_{\mathrm{1}} =−\mathrm{1}+\mathrm{2i}\:\mathrm{and}\:\mathrm{t}_{\mathrm{2}} =−\mathrm{1}−\mathrm{2i} \\ $$$$\mathrm{z}\:=−\mathrm{iln}\left(\mathrm{t}\right)\:\Rightarrow\mathrm{z}\:=−\mathrm{iln}\left(−\mathrm{1}\:\overset{−} {+}\mathrm{2i}\right) \\ $$$$ \\ $$
Commented by mr 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
$${but}\:{t}_{\mathrm{1}} =−\mathrm{1}+\mathrm{2}{i}={e}^{\left(\pi−\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\right){i}} \\ $$$${z}=−{i}\mathrm{ln}\:{t}_{\mathrm{1}} =\mathrm{ln}\:{t}_{\mathrm{1}} ^{−{i}} =\mathrm{ln}\:{e}^{\left(\pi−\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\right){i}\left(−{i}\right)} \\ $$$$=\pi−\mathrm{tan}^{−\mathrm{1}} \mathrm{2} \\ $$
Commented by mathmax by abdo last updated on 16/Jul/20
thank you sir i havent verify..
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{i}\:\mathrm{havent}\:\mathrm{verify}.. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *