Question

Find the value of $r$: ${\left(\frac{1+\cos \varphi +i\sin \varphi }{1+\cos \varphi -i\sin \varphi }\right)}^n=r(\cos n\varphi +i\sin n\varphi )$

Answer 1

$r=2$

${\left(\frac{1+\cos \varphi +i\sin \varphi }{1+\cos \varphi -i\sin \varphi }\right)}^n=r(\cos n\varphi +i\sin n\varphi )$

$\Rightarrow {\left[\frac{1+\cos \varphi +i\sin \varphi \left)\right(1+\cos \varphi +i\sin \varphi )}{(1+\cos \varphi -i\sin \varphi )(1+\cos \varphi +i\sin \varphi )}\right]}^n=r(\cos n\varphi +i\sin n\varphi )$

$\Rightarrow {\left[\frac{(1+\cos \varphi {)}^2-{\sin }^2\varphi +2i(1+\cos \varphi )\sin \varphi }{(1+\cos \varphi {)}^2+{\sin }^2\varphi }\right]}^n=r(\cos n\varphi +i\sin n\varphi )$

$\Rightarrow {\left[\frac{(1+\cos \varphi {)}^2-(1-{\cos }^2\varphi )+2i(1+\cos \varphi )\sin \varphi }{1+1+2\cos \varphi }\right]}^n=r(\cos n\varphi +i\sin n\varphi )$

$\Rightarrow {\left[\frac{(1+\cos \varphi {)}^2-(1-\cos \varphi \left)\right(1+\cos \varphi )+2i(1+\cos \varphi )\sin \varphi }{2(1+\cos \varphi )}\right]}^n=r(\cos n\varphi +i\sin n\varphi )$

$\Rightarrow {\left[\frac{(1+\cos \varphi )-(1-\cos \varphi )+2i\sin \varphi }{2}\right]}^n=r(\cos n\varphi +i\sin n\varphi )$

$\Rightarrow {\left[\frac{2\cos \varphi +2i\sin \varphi }{2}\right]}^n=r(\cos n\varphi +i\sin n\varphi )$

$\Rightarrow [\cos \varphi +i\sin \varphi {]}^n=r(\cos n\varphi +i\sin n\varphi )$

use De' maywer theorem -

$(\cos \varphi +i\sin \varphi {)}^n=\cos n\varphi +i\sin n\varphi$

So, $(\cos n\varphi +i\sin n\varphi )=r(\cos n\varphi +i\sin n\varphi )$

compare both sides

So, $r=1$