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Question: If \(z = 1\) and \(z_{1}\) are two non-zero complex numbers such that \(z_{2}\)then arg \(|z_{1} - z...

If z=1z = 1 and z1z_{1} are two non-zero complex numbers such that z2z_{2}then arg ∣z1−z2∣|z_{1} - z_{2}|arg ≥∣z1∣−∣z2∣\geq |z_{1}| - |z_{2}| is equal to.

A

≤∣z1∣−∣z2∣\leq |z_{1}| - |z_{2}|

B

≥∣z1∣+∣z2∣\geq |z_{1}| + |z_{2}|

C

≤∣z2∣−∣z1∣\leq |z_{2}| - |z_{1}|

D

0

Answer

0

Explanation

Solution

Let =1−3−2i31+3=−2−2i34=−12−i32arg(z)=tan⁡−1yx=tan⁡−13=π3=60o.= \frac{1 - 3 - 2i\sqrt{3}}{1 + 3} = \frac{- 2 - 2i\sqrt{3}}{4} = - \frac{1}{2} - i\frac{\sqrt{3}}{2}arg(z) = \tan^{- 1}\frac{y}{x} = \tan^{- 1}\sqrt{3} = \frac{\pi}{3} = 60^{o.},∵\because ar⥂g(z)ar ⥂ g(z)

180o60o=240o180^{o}60^{o} = 240^{o}

arg⁡(1−i31+i3)=arg(1−i3)−arg(1+i3)\arg\left( \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}} \right) = arg(1 - i\sqrt{3}) - arg(1 + i\sqrt{3})

=−60o−60o=−120o240o= - 60^{o} - 60^{o} = - 120^{o}240^{o}

Therefore z=sin⁡α+i(1−cos⁡α)z = \sin\alpha + i(1 - \cos\alpha)

Thus arg amp(z)=tan⁡−1(1−cos⁡αsin⁡α)=tan⁡−1(2sin⁡2α22sin⁡α2cos⁡α2)amp(z) = \tan^{- 1}\left( \frac{1 - \cos\alpha}{\sin\alpha} \right) = \tan^{- 1}\left( \frac{2\sin^{2}\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}} \right).

Trick : =tan⁡−1tan⁡(α2)=α2= \tan^{- 1}{\tan\left( \frac{\alpha}{2} \right)} = \frac{\alpha}{2}lies on same straight line.

arg(z)arg(z)