Convert to Trigonometric Form -1+i square root of 3

Problem

$-1+i\sqrt{3}$

Solution

1. Identify the real part $a=-1$ and the imaginary part $b=\sqrt{3}$ of the complex number $z=a+bi$

2. Calculate the modulus $r$ using the formula $r=\sqrt{a^2+b^2}$

$r=\sqrt{{(-1)}^2+{\left(\sqrt{3}\right)}^2}$

$r=\sqrt{1+3}$

$r=2$

3. Determine the argument $\theta$ by finding the angle such that $\cos \theta =\frac{a}{r}$ and $\sin \theta =\frac{b}{r}$

$\cos \theta =-\frac{1}{2}$

$\sin \theta =\frac{\sqrt{3}}{2}$

4. Locate the quadrant of the complex number. Since $a<0$ and $b>0$ the point $(-1,\sqrt{3})$ lies in the second quadrant.

$\theta =\frac{2\pi }{3}$

5. Substitute the values of $r$ and $\theta$ into the trigonometric form $z=r(\cos \theta +i\sin \theta )$

Final Answer

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

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