Obtain geometric and polar form of (2+√3i)^-1

plz help

(2+√3i)⁻¹
= 1 / (2+√3i)
= (2−√3i) / ((2+√3i)(2−√3i))
= (2−√3i) / (4 − 3)
= 2 − √3i

Polar form
r = √(4+3) = √7
tanθ = −√3/2
θ = arctan(−√3/2) = −40.89°

Be careful when calculating θ, since arctan always returns value in quadrant 4 or quadrant 1. Since 2 − √3i is in Q4, we are ok taking arctan(−√3/2), but for values in Q2 or Q3 you must add 180° to arctan.

2 − √3i = √7 (cos(−40.89°) + i sin(−40.89°))