Tìm \(\sqrt 4\) trong trường hợp số phức 

Ta có:
\(z = (1 + i\sqrt 3 )(1 - i) = 1 + \sqrt 3 + i(\sqrt 3 - 1)\)
Suy ra:
\(\tan \varphi = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} = \frac{{(\sqrt 3 - 1)(\sqrt 3 - 1)}}{{(\sqrt 3 + 1)(\sqrt 3 - 1)}} = \frac{{4 - 2\sqrt 3 }}{2} = 2 - \sqrt 3 \)
Vậy \(\varphi = \arctan (2 - \sqrt 3 ) = \frac{\pi }{{12}}\)

Ta có:

\(z = \frac{{ - 1 + i\sqrt 3 }}{{{{(1 + i)}^{15}}}} = \frac{{2(\cos \frac{{2\pi }}{3} + i\sin \frac{{2\pi }}{3})}}{{{{(1 + i)}^{15}}}}\)

Lại có:

\(1 + i = \sqrt 2 (\cos \frac{\pi }{4} + i\sin \frac{\pi }{4})\)

Do đó:

\({(1 + i)^{15}} = {(\sqrt 2 )^{15}}(\cos \frac{{15\pi }}{4} + i\sin \frac{{15\pi }}{4}) = {2^{\frac{{15}}{2}}}(\cos \frac{{7\pi }}{4} + i\sin \frac{{7\pi }}{4})\)

Vậy:

\(z = \frac{{2(\cos \frac{{2\pi }}{3} + i\sin \frac{{2\pi }}{3})}}{{{2^{\frac{{15}}{2}}}(\cos \frac{{7\pi }}{4} + i\sin \frac{{7\pi }}{4})}} = {2^{ - \frac{{13}}{2}}}(\cos (\frac{{2\pi }}{3} - \frac{{7\pi }}{4}) + i\sin (\frac{{2\pi }}{3} - \frac{{7\pi }}{4}))\)

\(z = {2^{ - \frac{{13}}{2}}}(\cos (\frac{{ - 13\pi }}{{12}}) + i\sin (\frac{{ - 13\pi }}{{12}}))\)

Vì argument của số phức có tính chất cộng \(2k\pi \), nên:

\(\varphi = - \frac{{13\pi }}{{12}} + 2\pi = \frac{{11\pi }}{{12}}\)