The argument of the complex number frac{13 - | vedclass

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Question

(B) Let $z = \frac{13 - 5i}{4 - 9i}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(4 + 9i)$:$z = \frac{(13

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$\frac{\pi}{4}$

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(B) Let $z = \frac{13 - 5i}{4 - 9i}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(4 + 9i)$:$z = \frac{(13 - 5i)(4 + 9i)}{(4 - 9i)(4 + 9i)}$$z = \frac{52 + 117i - 20i - 45i^2}{16 + 81}$$z = \frac{52 + 97i + 45}{97} = \frac{97 + 97i}{97} = 1 + i$The argument of $z = 1 + i$ is $	heta = 	an^{-1}(\frac{1}{1}) = \frac{\pi}{4}$.

The argument of the complex number $\frac{13 - 5i}{4 - 9i}$ is

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