Write the complex number in the form a + bi. 4[cos (-135°) + i sin (-135°)].

Solution:

Given, 4[cos (-135°) + i sin (-135°)]

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

We know that,

sin(-135°) = -√2/2

cos(-135°) = -√2/2

Putting the values,

= 4(-√2/2 + i (-√2/2))

= 4(-√2/2 - i√2/2)

Taking out (-√2/2) as common,

= 4(-√2/2)[1 + i]

= -2√2(1 + i)

= -2√2 - i2√2

Therefore, the complex number in the form a + bi is -2√2 - i2√2

---

Write the complex number in the form a + bi. 4[cos (-135°) + i sin (-135°)].

Summary:

The complex number 4(cos (-135°) + i sin (-135°)) in the form a + bi is -2√2 - i2√2.