If z is a complex number such that |z - bar{z | vedclass

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(D) Let $z = x + iy$. Given $|z + \bar{z}| = 4$,we have $|2x| = 4$,which implies $x = \pm 2$. Given $|z - \bar{z}| = 2$,we have $|2iy| = 2$,which impl

Vedclass · Accepted Answer

$Amp(z) \in (\frac{\pi}{6}, \frac{\pi}{4})$

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(D) Let $z = x + iy$. Given $|z + \bar{z}| = 4$,we have $|2x| = 4$,which implies $x = \pm 2$. Given $|z - \bar{z}| = 2$,we have $|2iy| = 2$,which implies $|y| = 1$,so $y = \pm 1$. The possible values for $z$ are $2+i, 2-i, -2+i, -2-i$. The argument $	heta = Amp(z)$ satisfies $	an 	heta = \frac{y}{x} = \pm \frac{1}{2}$. For $z = 2+i$,$	an 	heta = 1/2 \Rightarrow 	heta \in (0, \pi/6)$. For $z = 2-i$,$	an 	heta = -1/2 \Rightarrow 	heta \in (-\pi/6, 0)$. For $z = -2+i$,$	an 	heta = -1/2 \Rightarrow 	heta \in (5\pi/6, \pi)$. For $z = -2-i$,$	an 	heta = 1/2 \Rightarrow 	heta \in (-\pi, -5\pi/6)$. Comparing these with the options,the range $(\pi/6, \pi/4)$ is never attained by $Amp(z)$.

List-$I$ (Complex number)	List-$II$ (Polar form)
$(i) \sqrt{3}-i$	$(a) 2 \operatorname{cis} \frac{\pi}{6}$
$(ii) \sqrt{3}+i$	$(b) 2 \operatorname{cis} \frac{5 \pi}{6}$
$(iii) -\sqrt{3}+i$	$(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$
$(iv) -\sqrt{3}-i$	$(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$

If $z$ is a complex number such that $|z - \bar{z}| = 2$ and $|z + \bar{z}| = 4$,then which of the following is always incorrect -

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