Question

If |z| = 4 and arg(z) = `(5pi)/6`, then z = ______.

Answer 1

If |z| = 4 and arg(z) = `(5pi)/6`, then z = `underlinebb(-2 sqrt(3) + 2i)`.

Explanation:

Given that:  |z| = 4 and arg(z) = `(5pi)/6`

Let z = x + yi

|z| = `sqrt(x^2 + y^2)` = 4

⇒ x² + y² = 16   ......(i)

arg(z) = `tan^-1  (y/x) = (5pi)/6`

⇒ `y/x = tan  (5pi)/6`

= `tan(pi - pi/6)`

= `- tan  pi/6`

= `-1/sqrt(3)`

∴ x = `- sqrt(3) y`  ....(ii)

From equation (i) and (ii),

`(- sqrt(3) y)^2 + y^2` = 16

⇒ 3y² + y² = 16

⇒ 4y² = 16

⇒ y² = 4

⇒ y = `+-  2`

∴ x = `-2 sqrt(3)`

So, z = `-2 sqrt(3) + 2i`