# How do you convert -sqrt3+i  to polar form?

May 1, 2018

$2 \left[\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right)\right]$ or $2 c i s \frac{5 \pi}{6}$

#### Explanation:

Currently, $- \sqrt{3} + i$ is written in standard/rectangular form, or $a + b i$. To convert it into polar form, we must write it in the format $r \left(\cos \theta + i \sin \theta\right)$, where $r$ is the radius.

The radius is like the hypotenuse in a right triangle. You solve it like this:
$r = \sqrt{{a}^{2} + {b}^{2}}$

In this case, we know that $a = - \sqrt{3}$ and $b = 1$, so we can substitute them into the formula:
$r = \sqrt{{\left(- \sqrt{3}\right)}^{2} + {\left(1\right)}^{2}}$

Now simplify:
$r = \sqrt{3 + 1}$

$r = \sqrt{4}$

$r = 2$

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To solve for $\theta$, we use $\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$.

Let's substitute our values into the formula again:
$\theta = {\tan}^{-} 1 \left(\frac{1}{-} \sqrt{3}\right)$

$\theta = {\tan}^{-} 1 \left(- \frac{1}{\sqrt{3}}\right)$

$\theta = \frac{5 \pi}{6}$ or $\frac{11 \pi}{6}$

However, to determine the angle, we need to look at the graph of our expression $- \sqrt{3} + i$

Here's the graph for it. The real axis is the $x$-axis and the imaginary axis is the $y$-axis. In the graph, the $\textcolor{red}{\text{red dot}}$ is at $- \sqrt{3} + i$.

As we can see, the complex number is in the 2nd quadrant. Looking back at our angle, we know that $\theta$ is either $\frac{5 \pi}{6}$ or $\frac{11 \pi}{6}$. Since $\frac{5 \pi}{6}$ is in the second quadrant, we know that $\frac{5 \pi}{6}$ is $\theta$.

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Finally, we write this in polar form.

We know that $r = 2$ and $\theta = \frac{5 \pi}{6}$, so let's substitute it into polar form:
$\quad \quad \quad r \left(\cos \theta + i \sin \theta\right)$

$\implies 2 \left[\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right)\right]$

*The $\left[\right]$ brackets means the same as parenthesis, it's just for easier read.

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Instead of writing the long $2 \left[\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right)\right]$, you can also write it as a shortcut, as shown here:
$2 c i s \frac{5 \pi}{6}$

Where "$c i s$" refers to "$\cos i \sin$"

However, this is only if you teacher allows it ! If not, keep it as $2 \left[\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right)\right]$!

Hope this helps!