Question 4b(ii)

SCH_CALC

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$$\boxed{\text{Given that }z=\cos \theta +i\sin \theta ,\text{prove that: }2\text{arg}(z+1)=\text{arg}(z)}$$ $$\text{arg}\left[\cos \theta +i\sin \theta \right]={\tan }^{-1}\left[\frac{\sin \theta }{\cos \theta }\right]$$ $$={\tan }^{-1}\tan \theta =\theta$$ $$RHS=\theta$$ $$LHS=2\text{arg}(z+1)$$ $$LHS=2\left[{\tan }^{-1}\left(\frac{\sin \theta }{\cos \theta +1}\right)\right]$$

Maybe if the argument of the inverse tan be $\tan \frac{1}{2}\theta$ then it will cancel LHS to be $\theta$, then expression could be QED’ed?
Goal: re-write $\frac{\sin \theta }{\cos \theta +1}$ into a $\tan \theta$… form!

Multiplying by denominator’s conjugate:

$$LHS=2{\tan }^{-1}\left(\frac{\sin \theta (\cos \theta -1)}{{\cos }^2\theta +1}\right)$$ $$=2{\tan }^{-1}\left(\frac{\sin \theta (\cos \theta -1)}{1-{\sin }^2\theta +1}\right)$$ $$=2{\tan }^{-1}\left(\frac{\sin \theta (\cos \theta -1)}{-{\sin }^2\theta }\right)$$ $$=2{\tan }^{-1}\left(\frac{-\cos \theta +1}{\sin \theta }\right)$$ $$=2{\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)$$

Using the half angle identity of $\tan \frac{\theta }{2}=\frac{1-\cos \theta }{\sin \theta }$,

$$⟹LHS=2{\tan }^{-1}\left(\tan \left(\frac{\theta }{2}\right)\right)$$ $$=2\cdot \frac{\theta }{2}$$ $$=\theta =RHS∎$$