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Question: If \(z_{1} + z_{2}\) and \(\tan^{- 1}\left( \frac{2k}{k^{2} + 1} \right)\)then z =...

If z1+z2z_{1} + z_{2} and tan⁡−1(2kk2+1)\tan^{- 1}\left( \frac{2k}{k^{2} + 1} \right)then z =

A

tan⁡−1(2k1−k2)\tan^{- 1}\left( \frac{2k}{1 - k^{2}} \right)

B

2tan⁡−1k2\tan^{- 1}k

C

2tan⁡−1k2\tan^{- 1}k

D

∣z+1z∣=1\left| z + \frac{1}{z} \right| = 1

Answer

2tan⁡−1k2\tan^{- 1}k

Explanation

Solution

cos⁡(θ1−θ2)=0⇒θ1−θ2=π2\cos(\theta_{1} - \theta_{2}) = 0 \Rightarrow \theta_{1} - \theta_{2} = \frac{\pi}{2}and arg⁡(z1z2)=π2⇒Re(z1z2)=∣z1∣∣z2∣cos⁡(π2)=0\arg\left( \frac{z_{1}}{z_{2}} \right) = \frac{\pi}{2} \Rightarrow {Re}\left( \frac{z_{1}}{z_{2}} \right) = \frac{|z_{1}|}{|z_{2}|}\cos\left( \frac{\pi}{2} \right) = 0

Let Re(z1z2)=0⇒Re(z1z2‾)=0{Re}\left( \frac{z_{1}}{z_{2}} \right) = 0 \Rightarrow {Re}(z_{1}\overline{z_{2}}) = 0, then z1z2‾z_{1}\overline{z_{2}}

and ∣z1−z2∣2|z_{1} - z_{2}|^{2}

=∣z1∣2+∣z2∣2−2∣z1∣∣z2∣cos⁡(θ1−θ2)= |z_{1}|^{2} + |z_{2}|^{2} - 2|z_{1}||z_{2}|\cos(\theta_{1} - \theta_{2}) θ1=arg(z1)\theta_{1} = arg(z_{1}).

and θ2=arg(z2)∵\theta_{2} = arg(z_{2})\because

arg⁡z1−argz2=0\arg z_{1} - argz_{2} = 0 ∣z1−z2∣2=∣z1∣2+∣z2∣2−2∣z1∣∣z2∣=(∣z1∣−∣z2∣)2|z_{1} - z_{2}|^{2} = |z_{1}|^{2} + |z_{2}|^{2} - 2|z_{1}||z_{2}| = (|z_{1}| - |z_{2}|)^{2}

Trick : Since ∣z1−z2∣=∣∣z1∣−∣z2∣∣|z_{1} - z_{2}| = ||z_{1}| - |z_{2}||, here the complex number must lie in second quadrant, so (1) and (2) rejected. Also x1x2+y1y2=0x_{1}x_{2} + y_{1}y_{2} = 0 which satisfies (3) only.