How do you write the complex number in trigonometric form $- 2 (1 + \sqrt{3} i)$ $-2(1+sqrt3i)$ ?

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Answer 1 · 2017-02-23T07:58:21

The trigonometric form is $= - 4 (cos (\frac{π}{3}) + i sin (\frac{π}{3}))$ $=-4(cos(pi/3)+isin(pi/3))$

The trigonometric form of a complex number $z = a + i b$ $z=a+ib$ is

$z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$

Here, we have

$z = - 2 (1 + \sqrt{3} i)$ $z=-2(1+sqrt3i)$

The modulus is

$| z | = - 2 \sqrt{1 + 3} = - 2 \cdot 2$ $|z|=-2sqrt(1+3)=-2*2$

$z = - 4 (\frac{1}{2} + \frac{\sqrt{3}}{2} i)$ $z=-4(1/2+sqrt3/2i)$

$cos θ = \frac{1}{2}$ $costheta=1/2$ and $sin θ = \frac{\sqrt{3}}{2}$ $sintheta=sqrt3/2$

$θ = \frac{π}{3}$ $theta=pi/3$

Therefore,

$z = - 4 (cos (\frac{π}{3}) + i sin (\frac{π}{3}))$ $z=-4(cos(pi/3)+isin(pi/3))$

$= - 4 e^{\frac{π}{3} i}$ $=-4e^(pi/3i)$

How do you write the complex number in trigonometric form −2(1+√3i)-2(1+sqrt3i)?