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:

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

$theta ~~ 2.39$

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

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