Question

Convert in polar form   -1-i

Answer 1

$\huge{Hey Mate!!!}$

☆☞ Here is ur answer ☜☆

☆☞ Let z = -1-i. Then,

               r = l z l = √ (-1)2 + (-1)2 = √2

 

Let tan α = | Im (z) / Re (z) |. Then,

      tan α = | -1/ -1 | = 1 or α = π/4

Since the point ( -1, -1) representing z lies in the third quadrant. Therefore, the argument of z is given by 

                   θ = - (π - α) = - ( π - π/4) = -3π/4

So, the polar form of z = -1-i is

                  z = r ( cosθ + i sin θ )

                     = √2{ cos (-3π/4) + i sin (-3π/4) }

                     = √2 ( cos 3π/4 - i sin 3π/ 4) is the required answer

Hope it helps.

CHEERS!!!

Answer 2

Let z = -1-i. Then,

r = l z l = √ (-1)2 + (-1)2 = √2

Let tan α = | Im (z) / Re (z) |. Then,

tan α = | -1/ -1 | = 1 or α = π/4
Since the point ( -1, -1) representing z lies in the third quadrant. Therefore, the argument of z is given by

θ = - (π - α) = - ( π - π/4) = -3π/4

So, the polar form of z = -1-i is

z = r ( cosθ + i sin θ )

= √2{ cos (-3π/4) + i sin (-3π/4) }

= √2 ( cos 3π/4 - i sin 3π/ 4)

HOPE IT HELPS ✌