Argument of the complex number z = frac{13-5i | vedclass

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Question

(A) To find the argument of $z = \frac{13-5i}{4-9i}$,first simplify the complex number by multiplying the numerator and denominator by the conjugate o

Vedclass · Accepted Answer

$\frac{\pi}{4}$

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(A) To find the argument of $z = \frac{13-5i}{4-9i}$,first simplify the complex number by multiplying 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 - 81i^2}$Since $i^2 = -1$,we have:$z = \frac{52 + 97i + 45}{16 + 81} = \frac{97 + 97i}{97} = 1 + i$Now,the complex number is in the form $z = x + iy$ where $x = 1$ and $y = 1$.The argument $	heta$ is given by $	an^{-1}(\frac{y}{x})$:$	heta = 	an^{-1}(\frac{1}{1}) = 	an^{-1}(1) = \frac{\pi}{4}$Thus,the argument is $\frac{\pi}{4}$.

Argument of the complex number $z = \frac{13-5i}{4-9i}$,where $i = \sqrt{-1}$,is

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