How do you convert $z = i - \frac{1}{cos (\frac{π}{3})} + i sin (\frac{π}{3})$ $z= i -1/cos (pi/3) + isin(pi/3)$ in polar form?

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Answer 1 · 2017-06-04T20:41:04

Given: $z = i - \frac{1}{cos (\frac{π}{3})} + i sin (\frac{π}{3})$ $z= i -1/cos (pi/3) + isin(pi/3)$

We know that $- \frac{1}{cos (\frac{π}{3})} = - 2$ $-1/cos(pi/3) = -2$

$z = - 2 + i + i sin (\frac{π}{3})$ $z= -2+ i + isin(pi/3)$

We know that $sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $sin(pi/3) = sqrt3/2$

$z = - 2 + i (1 + \frac{\sqrt{3}}{2})$ $z= -2 + i(1 + sqrt3/2)$

$r = \sqrt{{(- 2)}^{2} + {(1 + \frac{\sqrt{3}}{2})}^{2}}$ $r = sqrt((-2)^2 + (1 + sqrt3/2)^2)$

$r \approx 2.7$ $r ~~ 2.7$

The angle is in the second quadrant:

$θ = π + {tan}^{- 1} (- \frac{1}{2} - \frac{\sqrt{3}}{4})$ $theta = pi + tan^-1(-1/2-sqrt3/4)$

$θ \approx 2.39$ $theta ~~ 2.39$

$z = 2.7 e^{i 2.39}$ $z = 2.7e^(i2.39)$

How do you convert z=i−1cos(π3)+isin(π3)z= i -1/cos (pi/3) + isin(pi/3) in polar form?