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How do you convert #-6 + j3# to polar form?

1 Answer

sankarankalyanam

Jun 24, 2018

#color(orange)("Polar form is "(6.7082, 153.43^@#

Explanation:

#z= -6 + j 3, " also " = r, theta#

#r = sqrt((-6)^2 + 3^2) = sqrt45 = 6.7082#

#theta = tan ^ (-1) (3 / -6) = tan ^ (-1) (-0.5) = -26.57^@ = 153.43^@, " point lies in II quadrant"#

#color(orange)("Polar form is "(6.7082, 153.43^@#

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