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

$$whatarethecomplexsolution\tan$$$$\left(z\right)=-2?$$

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

$$z=k\pi -{\tan }^{-1}2$$$$zisalwaysreal!$$

Commented by bemath last updated on 15/Jul/20

$$nothingincomplexsir?$$

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

$$arealnumberisalsoacomplex$$$$number.ifyouwant,$$$$z=k\pi -{\tan }^{-1}2+0i$$

Commented by bobhans last updated on 15/Jul/20

$$z=k\pi -{\tan }^{-1}\left(2\right)orz=k\pi -{\tan }^{-1}(-2)$$$$sir?$$

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

$$z=k\pi -{\tan }^{-1}\left(2\right)$$

Commented by bramlex last updated on 15/Jul/20

$$\tan \left(\mathrm{z}\right)=-2\Rightarrow \tan (-\mathrm{z})=2$$$$\mathrm{note}\tan \mathrm{x}=\tan \theta \Rightarrow \mathrm{x}=\theta +\mathrm{n}\pi$$$$\Rightarrow \tan (-\mathrm{z})=\tan \left({\tan }^{-1}\left(2\right)\right)$$$$-\mathrm{z}+\mathrm{n}\pi ={\tan }^{-1}\left(2\right)$$$$\mathrm{z}=\mathrm{n}\pi -{\tan }^{-1}\left(2\right)(\oplus )$$

Answered by OlafThorendsen last updated on 15/Jul/20

$$\tan z=-2$$$$\frac{\sin z}{\cos z}=-2$$$$\frac{\frac{e^{iz}-e^{-iz}}{2i}}{\frac{e^{iz}+e^{-iz}}{2}}=-2$$$$\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}=-2i$$$$e^{iz}-e^{-iz}=-2i(e^{iz}+e^{-iz})$$$$(1+2i)e^{iz}=(1-2i)e^{-iz}$$$$(1+2i)e^{2iz}=(1-2i)$$$$e^{2iz}=\frac{1-2i}{1+2i}=\frac{{(1-2i)}^2}{5}=-\frac{3+4i}{5}$$$$e^{2iz}=e^{i(\mathrm{Arctan}\frac{4}{3}+2k\pi )}$$$$z=\frac{1}{2}\mathrm{Arctan}\frac{4}{3}+k\pi ,k\in \mathbb{Z}$$$$\mathrm{Note}:$$$$(\frac{1}{2}\mathrm{Arctan}\frac{4}{3}-\pi =\mathrm{Arctan}(-2))$$

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

$$\mathrm{tanz}=-2\Rightarrow \frac{\mathrm{sinz}}{\mathrm{cosz}}=-2\Rightarrow \mathrm{sinz}=-2\mathrm{cosz}\Rightarrow$$$${\sin }^2\mathrm{z}={4\cos }^2\mathrm{z}\Rightarrow 1-{\cos }^2\mathrm{z}={4\cos }^2\mathrm{z}\Rightarrow 1={5\cos }^2\mathrm{z}\Rightarrow$$$${\cos }^2\mathrm{z}=\frac{1}{5}\Rightarrow \mathrm{cosz}=\stackrel{-}{+}\frac{1}{\sqrt{5}}$$$$\mathrm{cosz}=\frac{1}{\sqrt{5}}\Rightarrow \mathrm{ch}\left(\mathrm{iz}\right)=\frac{1}{\sqrt{5}}\Rightarrow \frac{{\mathrm{e}}^{\mathrm{iz}}+{\mathrm{e}}^{-\mathrm{iz}}}{2}=\frac{1}{\sqrt{5}}\Rightarrow {\mathrm{e}}^{\mathrm{iz}}+{\mathrm{e}}^{-\mathrm{iz}}=\frac{2}{\sqrt{5}}$$$${\mathrm{e}}^{\mathrm{iz}}=\mathrm{t}\Rightarrow \mathrm{t}+{\mathrm{t}}^{-1}=\frac{2}{\sqrt{5}}\Rightarrow {\mathrm{t}}^2+1=\frac{2}{\sqrt{5}}\mathrm{t}\Rightarrow {\mathrm{t}}^2-\frac{2\mathrm{t}}{\sqrt{5}}+1=0\Rightarrow$$$$\sqrt{5}{\mathrm{t}}^2-2\mathrm{t}+\sqrt{5}=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1-5=-4\Rightarrow {\mathrm{t}}_1=1+2\mathrm{i}\mathrm{and}{\mathrm{t}}_2=1-2\mathrm{i}$$$${\mathrm{e}}^{\mathrm{iz}}=\mathrm{t}\Rightarrow \mathrm{iz}=\mathrm{lnt}\Rightarrow \mathrm{z}=-\mathrm{iln}\left(\mathrm{t}\right)\Rightarrow \mathrm{z}=-\mathrm{iln}\left(1\stackrel{-}{+}2\mathrm{i}\right)$$$$\mathrm{case}2\mathrm{cosz}=-\frac{1}{\sqrt{5}}\Rightarrow \frac{{\mathrm{e}}^{\mathrm{iz}}+{\mathrm{e}}^{-\mathrm{iz}}}{2}=-\frac{1}{\sqrt{5}}\Rightarrow$$$${\mathrm{e}}^{\mathrm{iz}}+{\mathrm{e}}^{-\mathrm{iz}}=\frac{-2}{\sqrt{5}}\Rightarrow \mathrm{t}+{\mathrm{t}}^{-1}=-\frac{2}{\sqrt{5}}(\mathrm{t}={\mathrm{e}}^{\mathrm{iz}})\Rightarrow$$$${\mathrm{t}}^2+1=-\frac{2}{\sqrt{5}}\mathrm{t}\Rightarrow {\mathrm{t}}^2+\frac{2\mathrm{t}}{\sqrt{5}}+1=0\Rightarrow \sqrt{5}{\mathrm{t}}^2+2\mathrm{t}+\sqrt{5}=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1-5=-4\Rightarrow$$$${\mathrm{t}}_1=-1+2\mathrm{i}\mathrm{and}{\mathrm{t}}_2=-1-2\mathrm{i}$$$$\mathrm{z}=-\mathrm{iln}\left(\mathrm{t}\right)\Rightarrow \mathrm{z}=-\mathrm{iln}(-1\stackrel{-}{+}2\mathrm{i})$$$$$$

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

$$but{t}_1=-1+2i=e^{(\pi -{\tan }^{-1}2)i}$$$$z=-i\ln t_1=\ln t_1^{-i}=\ln e^{(\pi -{\tan }^{-1}2)i(-i)}$$$$=\pi -{\tan }^{-1}2$$

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

$$\mathrm{thank}\mathrm{you}\mathrm{sir}\mathrm{i}\mathrm{havent}\mathrm{verify}..$$

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