It is given that complex numbers $z_{1}$ and $z_{2}$ satisfy $| z_{1} |$ = 2 and $| z_{2} |$ = 3. If the included angle of their corresponding vectors is $60^{o}$ the $∣ ∣ \frac{z_{1} + z_{2}}{z_{1} - z_{2}} ∣ ∣$ can be expressed as $\frac{\sqrt{x}}{7}$ where `x` is a natural number the `x` equals to.

Q. If complex numbers

z_{1}, z_{2}

are such that

| z_{1} | = \sqrt{2}, | z_{2} | = \sqrt{3} and | z_{1} + z_{2} | = \sqrt{5 - 2 \sqrt{3}}

, then the value of

| arg (z_{1}) - arg (z_{2}) |

Q. If

z_{1}

and

z_{2}

are two complex numbers, then prove that

| z_{1} | + | z_{2} | = ∣ ∣ ∣ \frac{z_{1} + z_{2}}{2} + \sqrt{z_{1} z_{2}} ∣ ∣ ∣ + ∣ ∣ ∣ \frac{z_{1} + z_{2}}{2} - \sqrt{z_{1} z_{2}} ∣ ∣ ∣

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Complex numbers z1 and z2 satisfy |z1|=2 and |z2|=3. If the included angle of z2 their corresponding vectors is 60o, then the value of 19∣∣∣z1−z2z1+z2∣∣∣2 is

The correct option is B 7Let θ be the angle between z1 and z2.θ=60∘ (given)|z1|=2 and |z2|=3 (given)19∣∣∣z1−z2z1+z2∣∣∣2=19[|z1|2+|z2|2−2|z1||z2|cosθ|z1|2+|z2|2+2|z1||z2|cosθ]=19[22+32−2×2×3×cos60∘22+32+2×2×3×cos60∘]=19[4+9−64+9+6]=19×719=7 [C]

Complex numbers $z_{1}$ and $z_{2}$ satisfy $| z_{1} | = 2$ and $| z_{2} | = 3$ . If the included angle of $z_{2}$ their corresponding vectors is 60 $^{o}$ , then the value of $19 {∣ ∣ ∣ \frac{z_{1} - z_{2}}{z_{1} + z_{2}} ∣ ∣ ∣}^{2}$ is