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Question Number 127610 by pticantor last updated on 31/Dec/20

$$\mathit{Z}=1+(1+i)\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$$$$\mathit{a}\mathit{r}\mathit{g}\left(\mathit{z}\right)=?$$

Answered by MJS_new last updated on 31/Dec/20

$$\arg (1+\cos \theta +\mathrm{i}\cos \theta )=$$$$=\frac{\pi }{2}\mathrm{sign}\left(\cos \theta \right)-\arctan \frac{1+\cos \theta }{\cos \theta }$$

Commented by pticantor last updated on 31/Dec/20

$$pls{can}^{\prime }tyouexplain?$$

Commented by MJS_new last updated on 31/Dec/20

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{just}\mathrm{the}\mathrm{formula}$$$$\arg (a+b\mathrm{i})=\frac{\pi }{2}\mathrm{sign}b-\arctan \frac{a}{b}$$$$\mathrm{which}\mathrm{gives}\mathrm{an}\mathrm{angle}\phi \mathrm{with}-\pi ⩽\phi <\pi$$$$\mathrm{test}\mathrm{it}\mathrm{for}z=a+b\mathrm{i}\mathrm{in}\mathrm{all}4\mathrm{quadrants}:$$$$z=\{\begin{array}{l}+1\\ -1\\ -1\\ +1\end{array}\begin{array}{c}+\mathrm{i}\\ +\mathrm{i}\\ -\mathrm{i}\\ -\mathrm{i}\end{array}\}\Rightarrow \arg z=\{\begin{array}{l}\pi /2-\arctan 1=\pi /4\\ \pi /2-\arctan (-1)=3\pi /4\\ -\pi /2-\arctan 1=-3\pi /4\\ -\pi /2-\arctan (-1)=-\pi /4\end{array}$$

Answered by mathmax by abdo last updated on 01/Jan/21

$$\mathrm{Z}=1+(1+\mathrm{i})\cos \theta =1+\cos \theta +\mathrm{icos}\theta \Rightarrow \mid \mathrm{Z}\mid =\sqrt{{(1+\cos \theta )}^2+{\cos }^2\theta }$$$$\Rightarrow \mathrm{Z}=\mid \mathrm{Z}\mid (\frac{1+\cos \theta }{\mid \mathrm{Z}\mid }+\mathrm{i}\frac{\cos \theta }{\mid \mathrm{Z}\mid })=\mid \mathrm{Z}\mid {\mathrm{e}}^{\mathrm{i}\alpha }\Rightarrow \cos \alpha =1+\cos \theta \mathrm{and}\sin \alpha =\cos \theta$$$$\Rightarrow \tan \alpha =\frac{\cos \theta }{1+\cos \theta }\Rightarrow \alpha =\arctan \left(\frac{\cos \theta }{1+\cos \theta }\right)=\arg \left(\mathrm{Z}\right)$$

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