What is the polar form of Negative 2 StartRoot 3 EndRoot minus 6 i?

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)]

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer