Convert to Trigonometric Form 1- square root of 3i

Problem

$1-\sqrt{3}i$

Solution

1. Identify the real part $a$ and the imaginary part $b$ of the complex number $z=a+bi$

$a=1$

$b=-\sqrt{3}$

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

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

$r=\sqrt{1+3}$

$r=2$

3. Determine the argument $\theta$ using the relation $\tan \theta =\frac{b}{a}$

$\tan \theta =\frac{-\sqrt{3}}{1}$

$\tan \theta =-\sqrt{3}$

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

$\theta ={300}^{\circ }$

$\theta =\frac{5\pi }{3}$

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

$z=2(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3})$

Final Answer

$1-\sqrt{3}i=2(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3})$

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