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Fill in the Blanks

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

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#### Solution

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

⇒ x^{2} + y^{2} = 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

⇒ 3y^{2} + y^{2} = 16

⇒ 4y^{2} = 16

⇒ y^{2} = 4

⇒ y = `+- 2`

∴ x = `-2 sqrt(3)`

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

Concept: Argand Plane and Polar Representation

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