How do you convert #z= i -1/cos (pi/3) + isin(pi/3)# in polar form?

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Answer 1 · 2017-06-04T20:41:04

Given: #z= i -1/cos (pi/3) + isin(pi/3)#

We know that #-1/cos(pi/3) = -2#

#z= -2+ i + isin(pi/3)#

We know that #sin(pi/3) = sqrt3/2#

#z= -2 + i(1 + sqrt3/2)#

#r = sqrt((-2)^2 + (1 + sqrt3/2)^2)#

#r ~~ 2.7 #

The angle is in the second quadrant:

#theta = pi + tan^-1(-1/2-sqrt3/4)#

#theta ~~ 2.39#

#z = 2.7e^(i2.39)#