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

Option A -
`z=√(3)-i`

Explanation:

Given : Polar representation of complex number
`z=2[\cos( (11\pi )/(6)) + i\sin( (11\pi )/(6) )]`

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

Solution :

`z=2[\cos((11\pi )/(6)) + i\sin((11\pi )/(6))]`

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

Where, r=2 and
`\theta=(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( (11\pi )/(6))+2\sin( (11\pi )/(6))\ i\implies 2\cdot \cfrac{√(3)}{2}+2\cdot \cfrac{-1}{2}\ i\implies √(3)-1i\\\\\boxed{z=√(3)-i}`

Therefore, Option A is correct.

Answer 2

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