Question

Find, Modulus and argument of z=-1-i



class 11 marks 4​

Answer 1

$let \: z = - 1 - i \\ |z| = \sqrt{( { - 1})^{2} +( { - 1})^{2} } = \sqrt{2} \\ \\ \theta = {tan}^{ - 1} \frac{ - 1}{ - 1} \\ = {tan}^{ - 1} 1 \\ = \frac{\pi}{4} \\ \\ since \: \: the \: \: point \: \: lies \: \\ \: in \: \: the \: \: third \: quadrent \\ arg( \theta) = - \pi + \theta \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - \pi + \frac{\pi}{4} \\ \: \: \: \: \: \: \: \: \: = \frac{ - 3\pi}{4}$