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Question: If\({Im}(z) = 0\)then \({Re}(z) = 0\)...

IfIm(z)=0{Im}(z) = 0then Re(z)=0{Re}(z) = 0

A

amp(z)=πamp(z) = \pi

B

∣z1+z12−z22∣|z_{1} + \sqrt{z_{1}^{2} - z_{2}^{2}}|

C

+∣z1−z12−z22∣+ |z_{1} - \sqrt{z_{1}^{2} - z_{2}^{2}}|

D

∣z1∣|z_{1}|

Answer

+∣z1−z12−z22∣+ |z_{1} - \sqrt{z_{1}^{2} - z_{2}^{2}}|

Explanation

Solution

If =tan⁡−1∞=π2= \tan^{- 1}\infty = \frac{\pi}{2}

arg⁡(z1z2)=π\arg\left( \frac{z_{1}}{z_{2}} \right) = \pi

Thus arg(z1)−arg(z2)=πarg(z_{1}) - arg(z_{2}) = \pi

Since the complex number lies in III quadrant, therefore

arg(z1)=arg(z2)+πarg(z_{1}) = arg(z_{2}) + \pi is

arg⁡\arg + (z1)=π+θ(z_{1}) = \pi + \theta

Aliter : z1=∣z1∣[cos⁡(π+θ)+isin⁡(π+θ)]z_{1} = |z_{1}|\lbrack\cos(\pi + \theta) + i\sin(\pi + \theta)\rbrack

=∣z1∣(−cos⁡θ−isin⁡θ)= |z_{1}|( - \cos\theta - i\sin\theta) or z2=∣z2∣(cos⁡θ+isin⁡θ)z_{2} = |z_{2}|(\cos\theta + i\sin\theta).