Write the complex number in the form a + bi. 6(cos 330° + i sin 330°)

Solution:

Given, 6[cos (330°) + i sin (330°)]

We have to write the complex number in the form a + bi.

We know that,

Sin (330°) = sin (360° - 30°) = -sin 30° = -1/2

cos (330°) = cos (360° - 30°) = cos 30° = √3/2

Since (360° - 30°) lies in the fourth quadrant, cos is positive and sin is negative.

Putting the values,

= 6(√3/2 + i (-1/2))

= 6(√3/2 - i 1/2)

Taking out (1/2) as common,

= 6(1/2)[√3 - i]

= 3(√3 - i)

= 3√3 - 3i

Therefore, the complex number in the form a + bi is 3√3 - 3i.

---

Write the complex number in the form a + bi. 6(cos 330° + i sin 330°)

Summary:

The complex number 6(cos (330°) + i sin (330°)) in the form a + bi is 3√3 - 3i.