I couldn't succeed converting the following complex number to it's trigonometric form, could someone help?
√(2+√3) + i*√(2-√3)
Put $\displaystyle z=\sqrt{2+\sqrt{3}}+i\sqrt{2-\sqrt{3}}\Longrightarrow |z|^2=2+\sqrt{3}+2-\sqrt{3}=4$,
$\displaystyle \arg(z)=\arctan\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\arctan\sqrt{7-4\sqrt{3}}=\frac{1}{12\pi}$ , so
$\displaystyle z=2cis(\frac{1}{12\pi})=2e^{i\frac{1}{12\pi}$
Tonio