I couldn't succeed converting the following complex number to it's trigonometric form, could someone help?

√(2+√3) + i*√(2-√3)

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- Nov 3rd 2010, 03:42 AMFujiwaraConverting to Trigonometric form
I couldn't succeed converting the following complex number to it's trigonometric form, could someone help?

√(2+√3) + i*√(2-√3) - Nov 3rd 2010, 04:49 AMtonio

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