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- the - z 1- z 2- z 1- z 2- can be expressed as -x-7 where - x - is a natural number the - x - equals to-A- 19B- 119C- 126D- 133

The correct option is D 133 |z_1+z_2| = √(|z_1|^2+|z_2|^2+2|z_1||z_2|cos 60^o) = √(19) |z_1 z_2| = √(|z_1|^2+|z_2|^2 2|z_1||z_2|cos 60^o) = √(7) | z_1+z_2/z_1 ...

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Byju's Answer

Standard XII

Physics

Antiderivative

It is given t...

Question

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.

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119

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126

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133

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Solution

The correct option is D
133

$| z_{1} + z_{2} |$ = $\sqrt{| z_{1} |^{2} + | z_{2} |^{2} + 2 | z_{1} | | z_{2} | c o s 60^{o}}$ = $\sqrt{19}$

$| z_{1} - z_{2} |$ = $\sqrt{| z_{1} |^{2} + | z_{2} |^{2} - 2 | z_{1} | | z_{2} | c o s 60^{o}}$ = $\sqrt{7}$

$∣ ∣ \frac{z_{1} + z_{2}}{z_{1} - z_{2}} ∣ ∣$ = $\sqrt{\frac{19}{7}}$ = $\frac{\sqrt{133}}{7}$

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It is given that complex numbers z1 and z2 satisfy |z1| = 2 and |z2| = 3. If the included angle of their corresponding vectors is 60o the ∣∣z1+z2z1−z2∣∣ can be expressed as √x7 where `x` is a natural number the `x` equals to.

The correct option is D 133 |z1+z2| = √|z1|2+|z2|2+2|z1||z2|cos60o = √19 |z1−z2| = √|z1|2+|z2|2−2|z1||z2|cos60o = √7 ∣∣z1+z2z1−z2∣∣ = √197 = √1337