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Question Number 54093 by Tawa1 last updated on 28/Jan/19

If tan(z) = 2, find the value of z

$$\mathrm{If}\:\:\:\:\:\:\mathrm{tan}\left(\mathrm{z}\right)\:\:=\:\:\mathrm{2},\:\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{z} \\ $$

Commented by mr W last updated on 28/Jan/19

z=tan^(−1) 2+nπ

$${z}=\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+{n}\pi \\ $$

Commented by Tawa1 last updated on 29/Jan/19

God bless you sir. But sir, z = x + iy, is it the same thing ?

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{But}\:\mathrm{sir},\:\:\:\mathrm{z}\:=\:\mathrm{x}\:+\:\mathrm{iy},\:\:\mathrm{is}\:\mathrm{it}\:\mathrm{the}\:\mathrm{same}\:\mathrm{thing}\:? \\ $$

Commented by mr W last updated on 29/Jan/19

z=x+y i tan z=((sin (2x)+i sinh (2y))/(cosh (2y)+cos (2x)))=2 sinh (2y)=0⇒y=0 sin (2x)=2 [cosh (2y)+cos (2x)] 2 sin x cos x=2 [1+2cos^2 x−1] sin x=2cos x tan x=2⇒x=tan^(−1) 2+nπ ⇒z=tan^(−1) 2+nπ+0i

$${z}={x}+{y}\:{i} \\ $$$$\mathrm{tan}\:{z}=\frac{\mathrm{sin}\:\left(\mathrm{2}{x}\right)+{i}\:\mathrm{sinh}\:\left(\mathrm{2}{y}\right)}{\mathrm{cosh}\:\left(\mathrm{2}{y}\right)+\mathrm{cos}\:\left(\mathrm{2}{x}\right)}=\mathrm{2} \\ $$$$\mathrm{sinh}\:\left(\mathrm{2}{y}\right)=\mathrm{0}\Rightarrow{y}=\mathrm{0} \\ $$$$\mathrm{sin}\:\left(\mathrm{2}{x}\right)=\mathrm{2}\:\left[\mathrm{cosh}\:\left(\mathrm{2}{y}\right)+\mathrm{cos}\:\left(\mathrm{2}{x}\right)\right] \\ $$$$\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}=\mathrm{2}\:\left[\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \:{x}−\mathrm{1}\right] \\ $$$$\mathrm{sin}\:{x}=\mathrm{2cos}\:{x} \\ $$$$\mathrm{tan}\:{x}=\mathrm{2}\Rightarrow{x}=\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+{n}\pi \\ $$$$\Rightarrow{z}=\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+{n}\pi+\mathrm{0}{i} \\ $$

Commented by Tawa1 last updated on 29/Jan/19

God bless you sir, i appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}.\: \\ $$

Commented by Tawa1 last updated on 29/Jan/19

Please sir. How is cosh(2y) = 1 sir from cosh(2y) + cos(2x)

$$\mathrm{Please}\:\mathrm{sir}.\:\:\mathrm{How}\:\mathrm{is}\:\mathrm{cosh}\left(\mathrm{2y}\right)\:=\:\mathrm{1}\:\:\mathrm{sir}\:\:\:\:\mathrm{from}\:\:\:\mathrm{cosh}\left(\mathrm{2y}\right)\:+\:\mathrm{cos}\left(\mathrm{2x}\right) \\ $$

Commented by Tawa1 last updated on 29/Jan/19

Ohh, i get. Since y = 0 right ?

$$\mathrm{Ohh},\:\mathrm{i}\:\mathrm{get}.\:\:\mathrm{Since}\:\:\mathrm{y}\:=\:\mathrm{0}\:\:\mathrm{right}\:? \\ $$

Commented by mr W last updated on 29/Jan/19

yes

$${yes} \\ $$

Commented by Tawa1 last updated on 29/Jan/19

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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