If the complex number z = 2 - i(2 tan frac{5 | vedclass

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(A) Given $z = 2 - i(2 	an \frac{5 \pi}{8})$.Comparing with $z = x + iy$,we have $x = 2$ and $y = -2 	an \frac{5 \pi}{8}$.The modulus $r$ is given b

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$(2 \sec \frac{3 \pi}{8}, \frac{3 \pi}{8})$

Vedclass · Answer

(A) Given $z = 2 - i(2 	an \frac{5 \pi}{8})$.Comparing with $z = x + iy$,we have $x = 2$ and $y = -2 	an \frac{5 \pi}{8}$.The modulus $r$ is given by $r = \sqrt{x^2 + y^2} = \sqrt{2^2 + (-2 	an \frac{5 \pi}{8})^2} = \sqrt{4(1 + 	an^2 \frac{5 \pi}{8})} = \sqrt{4 \sec^2 \frac{5 \pi}{8}} = |2 \sec \frac{5 \pi}{8}|$.Since $\frac{5 \pi}{8}$ is in the second quadrant,$\sec \frac{5 \pi}{8}$ is negative,so $r = -2 \sec \frac{5 \pi}{8} = 2 \sec(\pi - \frac{5 \pi}{8}) = 2 \sec \frac{3 \pi}{8}$.The argument $	heta$ is $	an^{-1}(\frac{y}{x}) = 	an^{-1}(\frac{-2 	an \frac{5 \pi}{8}}{2}) = 	an^{-1}(-	an \frac{5 \pi}{8}) = 	an^{-1}(	an(\pi - \frac{5 \pi}{8})) = 	an^{-1}(	an \frac{3 \pi}{8}) = \frac{3 \pi}{8}$.Thus,$(r, 	heta) = (2 \sec \frac{3 \pi}{8}, \frac{3 \pi}{8})$.

If the complex number $z = 2 - i(2 \tan \frac{5 \pi}{8})$ has modulus $r$ and argument $\theta$,then what are $(r, \theta)$?

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