Converting to Trigonometric form

**Fujiwara** · November 3rd 2010, 03:42 AM

I couldn't succeed converting the following complex number to it's trigonometric form, could someone help?
√(2+√3) + i*√(2-√3)

**tonio** · November 3rd 2010, 04:49 AM

Originally Posted by Fujiwara

I couldn't succeed converting the following complex number to it's trigonometric form, could someone help?
√(2+√3) + i*√(2-√3)

Put $z=\sqrt{2+\sqrt{3}}+i\sqrt{2-\sqrt{3}}\Longrightarrow |z|^2=2+\sqrt{3}+2-\sqrt{3}=4$ ,

$\arg(z)=\arctan\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\arctan\sqrt{7-4\sqrt{3}}=\frac{1}{12\pi}$ , so

$z=2cis(\frac{1}{12\pi})=2e^{i\frac{1}{12\pi}$

Tonio

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