Answer:
4√3 [cos (2π/3) + isin(2π/3)]
Step-by-step explanation:
We are to write out the polar form of:
-2√3 - 6i
In Mathematics:
The polar form of a given complex number a + bi is given as r[cos(θ)+isin(θ)],
where r =√a²+b²and
θ = arc tan (b/a) when a is > 0
or arc tan (b/a) + π or arc tan (b/a) + 180 when a is < 0
From the above question:
-2√3 - 6i
We have that
a = −2√3 and b = −6
Hence:
r =√a²+b²
r = √(-2√3)² + (-6)²
r = √(4 × 3) + 36
r = √12 + 36
r = √48
r = √16 × 3
r = 4√3
a is < 0
Hence,
θ = arc tan (b/a) + π
θ = arc tan (-6/-2√3) + π
θ = 2π/3
Substituting 4√3 for r and 2π/3 for θ in:
r[cos(θ)+isin(θ)]
-2√3 - 6i = 4√3 [cos (2π/3) + isin(2π/3)]
