Polar Form of a Complex Number

$x \space=\space \sqrt{3}-\imath$ $x \space = \space 2(\frac{\sqrt{3}}{2}-\imath\frac{1}{2})$ $x \space= \space 2(\cos{(2\pi-\frac{\pi}{6})-\imath\sin{(2\pi-\frac{\pi}{6}})})$ $x\space=\space2(\cos{\frac{11\pi}{6}}-\imath\sin{\frac{11\pi}{6}})$

Now $\theta \space = \space \frac{11\pi}{6}$ . So, answer 11+6= $\boxed{17}$

We have that $\large x=a+bi=\sqrt{3}-i \implies a=\sqrt{3},b=-1$

Converting the complex number to its polar form gives us:

$\large z=r(\cos(\theta)+i\sin(\theta))=\sqrt{(\sqrt{3})^2+(-1)^2}\left(\cos\left(\frac{11 \pi}{6}\right)+i\sin\left(\frac{11 \pi}{6}\right)\right)$ $\large = 4\left(\cos\left(\frac{11 \pi}{6}\right)+i\sin\left(\frac{11 \pi}{6}\right)\right)$

Therefore $\large a+b=11+6=\boxed{17}$ .

r = 2. So 3^(1/2) - i = 2 ( cos P + i sin P ) --> 3^(1/2)/2 - i/2 = cos P + i sin P. because cos P is positive and sin P is negative so the angle in 3th cuadrant arc cos 3^(1/2)/2 = (360 - 30) = 330. So 330 = 11/6 . 180 = 11/6 . phi ----> a + b = 11 + 6 = 17

first part is positive and second negative so cos is positive and sin is negative so theta ends in 4th quadrant so answer will be 11/6 or a+b = 17

the ratio of cos to sin is - \sqrt{3} which means tan theta is - \sqrt{3}. Which means theta is n*\pi - \frac{\pi}{6} . But since r is positive. It has to be 2\pi - \frac{\pi}{6} since \pi - \frac{\pi}{6} will give negative of what we want.

• X = √3 – i
• X = 2 ((√3/2) + i(-1/2))
• X = 2 (cos(11π/6) + isin(11π/6))
• Comparing with x = r[cosθ + isinθ]
• Θ = 11π/6
• Comparing with Θ = aπ/b; a = 11, b = 6
• Therefore, a + b = 17.

[I haven't really understand trig. I was also guessing the $r$ , but I think my solution is right]

From the problems, we have $x=r(\cos\theta+i\sin\theta)=\sqrt{3}-i$ , hence we also have

$\sqrt{3}-r\cos\theta=ri\sin\theta+i...(1)$

Like I said, I was guessing for the value of $r$ . And I get the possible value of $r$ is $2$ .

The right side of (1) is arbitrary, meanwhile its left side is real number. Therefore, it must follow that the right side is also real number. And the only possible real value of the right side of equation is

$2i\sin\theta+i=0...(2)$

Simplify this we get $\sin\theta=- \frac{1}{2}$ . One of the solution to the equation is $\theta=330^\circ$ .

Recall that $2i\sin\theta+i=0$ , thus $\sqrt{3}-2\cos\theta=0$ . We can use the last equation to verify (2).

$\cos\theta=\frac{\sqrt{3}}{2}\Leftrightarrow\theta=330^\circ$ [this is one of the values of $\theta$ ]

Hence, we conclude $\theta=330^\circ=\frac{11\pi}{6}$ . And finally $a+b=11+6=\boxed{17}$