Question

PRECALC QUESTION! Convert the polar representation of this complex number into its standard form: z = 2(cos 11pi/6+ i sin 11pi/6)? A. sqrt(3) -i B. 1- sqrt(3) i C. -sqrt(3)/ 2 -(1/2) i D. -sqrt(3) + i

Answer 1

$\bf z=2\left[ cos\left( \frac{11\pi }{6} \right) + i\ sin\left( \frac{11\pi }{6} \right) \right]\qquad \begin{cases} r=2\\ \theta=\frac{11\pi }{6}\\ -----\\ a=x=rcos(\theta)\\ b=y=rsin(\theta) \end{cases}\implies a+bi \\\\\\ 2cos\left( \frac{11\pi }{6} \right)+2sin\left( \frac{11\pi }{6} \right)\ i\implies 2\cdot \cfrac{\sqrt{3}}{2}+2\cdot \cfrac{-1}{2}\ i\implies \sqrt{3}-1i \\\\\\ \boxed{\sqrt{3}-i}$

Answer 2

Answer:

Option A - $z=\sqrt{3}-i$

Step-by-step explanation:

Given : Polar representation of complex number  $z=2[\cos( \frac{11\pi }{6}) + i\sin( \frac{11\pi }{6} )]$

To find : Convert the polar representation of this complex number into its standard form?

Solution :

$z=2[\cos(\frac{11\pi }{6}) + i\sin(\frac{11\pi }{6})]$

The given complex number is in the form, $z=r(\cos\theta+i\sin\theta)$

Where, r=2 and $\theta=\frac{11\pi }{6}$

The standard form of complex number is z=x+iy

Where, $a=x=r\cos\theta\\b=y=r\sin\theta$

$2\cos( \frac{11\pi }{6})+2\sin( \frac{11\pi }{6})\ i\implies 2\cdot \cfrac{\sqrt{3}}{2}+2\cdot \cfrac{-1}{2}\ i\implies \sqrt{3}-1i\\\\\boxed{z=\sqrt{3}-i}$  

Therefore, Option A is correct.