If z1=(3+i)2(1−3i)1+i and z2=(1+3i)2(3−i)1−i, then

z1=2eπi/622e−πi/32eπi/4=42e−πi/4z2=2eπj/322e−πi/62e−πi/4=42e2π3−π6+π4=42e3πi/4∴ z1|=|z2∣=42Amp⁡z1=−π/4,Amp⁡z2=3π/4∴ 3Amp⁡z1+Amp⁡z2=−3π/4+3π/4=0.

If z1=(3+i)2(1−3i)1+i and z2=(1+3i)2(3−i)1−i, then

z1=2eπi/622e−πi/32eπi/4=42e−πi/4z2=2eπj/322e−πi/62e−πi/4=42e2π3−π6+π4=42e3πi/4∴ z1|=|z2∣=42Amp⁡z1=−π/4,Amp⁡z2=3π/4∴ 3Amp⁡z1+Amp⁡z2=−3π/4+3π/4=0.

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If $z_{1} = \frac{(\sqrt{3} + i)^{2} (1 - \sqrt{3} i)}{1 + i} and z_{2} = \frac{(1 + \sqrt{3} i)^{2} (\sqrt{3} - i)}{1 - i}$ , then

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$\begin{aligned} amp z_{1} + amp z_{2} = 0 \end{aligned}$

$\begin{array}{r} 3 (amp z_{1}) + amp z_{2} = 0 \end{array}$

$\begin{array}{r} |z_{1}| = 2 |z_{2}| \end{array}$

$3 |z_{1}| = |z_{2}|$

answer is B.

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Detailed Solution

$\begin{array}{l} z_{1} = \frac{{(2 e^{π i / 6})}^{2} (2 e^{- π i / 3})}{\sqrt{2} e^{π i / 4}} = 4 \sqrt{2} e^{- π i / 4} \\ z_{2} = \frac{{(2 e^{π j / 3})}^{2} (2 e^{- π i / 6})}{\sqrt{2} e^{- π i / 4}} = 4 \sqrt{2} e^{\frac{2 π}{3} - \frac{π}{6} + \frac{π}{4}} \\ = 4 \sqrt{2} e^{3 π i / 4} \\ ∴ z_{1} | = | z_{2} ∣= 4 \sqrt{2} \end{array}$