My Coding > Mathematics > Complex numbers > Express 3 + i√3 in polar form

Express 3 + i√3 in polar form

Another classical task of expressing complex number \(3 + i\sqrt{3}\) in polar form. Again I will show you a straightforward way of calculating \(Arg(3, \sqrt{3})\) through simple trigonometric equations.

Let's start from calculating standard parameters for Eulers form:

\[r = \sqrt{3^2 + \sqrt{3}^2} = \sqrt{9+3} = \sqrt{12} = 2\sqrt{3}\]

Then we need to find \(Arg(3, \sqrt{3})\). I will find it via trigonometrical form of complex numbers \(z = r\times(\cos(\theta) + i\sin(\theta))\):

\[3 + i\sqrt{3} = 2\sqrt{3}\times(x + iy) \Rightarrow 2 = 2\sqrt{3}; \sqrt{3} = 2\sqrt{3}y \Rightarrow x = \frac{3}{2\sqrt{3}}; y = \frac{1}{2}\]

\[3 + i\sqrt{3} = 2\sqrt{3}(\frac{\sqrt{3}}{2} + i\frac{1}{2})\]

Comaring with the trigonometrical form of complex number, we can write:

\[\cos(\theta) = \frac{\sqrt{3}}{2}; \sin(\theta) = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6} + 2\pi n\]

therefore:

\[3 + i\sqrt{3} = 2\sqrt{3}\times(\cos(\frac{\pi}{6} + 2\pi n) + i\sin(\frac{\pi}{6} + 2\pi n)) \equiv \sqrt{2}\times e^{i(\frac{\pi}{6} + 2\pi n)}\]

with principal value:

\[3 + i\sqrt{3} = 2\sqrt{3}\times e^{i\frac{\pi}{6}}\]

For more detailed explanation, please go to watch our video:


Published: 2023-05-10 02:32:29
Updated: 2023-05-10 02:37:27

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