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Question: If \(\left( \frac{3 - 4ix}{3 + 4ix} \right) =\) then \(\alpha - i\beta(\alpha,\beta\text{real}),\) +...

If (3−4ix3+4ix)=\left( \frac{3 - 4ix}{3 + 4ix} \right) = then α−iβ(α,βreal),\alpha - i\beta(\alpha,\beta\text{real}), +α2−β2=−1\alpha^{2} - \beta^{2} = - 1 and α2+β2=1\alpha^{2} + \beta^{2} = 1 differ by a.

A

Multiple of α2−β2=2\alpha^{2} - \beta^{2} = 2

B

Multiple of z1z_{1}

C

Greater than ∣z1∣=1|z_{1}| = 1

D

Less than z2z_{2}

Answer

Multiple of α2−β2=2\alpha^{2} - \beta^{2} = 2

Explanation

Solution

We know that the principal value of arg⁡(13−5i4−9i)=arg(13−5i)−arg(4−9i)\arg\left( \frac{13 - 5i}{4 - 9i} \right) = arg(13 - 5i) - arg(4 - 9i) lies between =−tan⁡−1(513)+tan⁡−194=π4= - \tan^{- 1}\left( \frac{5}{13} \right) + \tan^{- 1}\frac{9}{4} = \frac{\pi}{4} and z1=r1(cos⁡θ1+isin⁡θ1)z_{1} = r_{1}(\cos\theta_{1} + i\sin\theta_{1}).