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Question
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
⇒ 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`
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