If for a complex number z 1 and z 2- z 1- z 2-0- then - z 1- z 2- is equal toA- - z 1- z 2-B- - z 1- z 2-C- z 1- z 2D- 0

We have, |z_1 z_2|^2 = |z_1|^2 + |z_2|^2 2Re(z_1z_2) ⇒ |z_1 z_2|^2 = |z_1|^2 + |z_2|^2 2Re(|z_1|e^i z_1|z_2|e^ i z_2) ⇒ |z_1 z_2|^2 = |z_1|^2 + |z_2|^2 2R ...

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Byju's Answer

Standard XII

Mathematics

Geometrical Representation of Argument and Modulus

If for a comp...

Question

If for a complex number $z_{1}$ and $z_{2}$ , $arg (z_{1}) - arg (z_{2}) = 0$ , then $| z_{1} - z_{2} |$ is equal to

| z_{1} | + | z_{2} |

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| z_{1} + z_{2} |

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| | z_{1} | - | z_{2} | |

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Solution

The correct option is C $| | z_{1} | - | z_{2} | |$
We have,
$| z_{1} - z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} - 2 R e (z_{1}_{2})$
$\Rightarrow | z_{1} - z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} - 2 R e (| z_{1} | e^{i arg z_{1}} |_{2} | e^{- i arg z_{2}})$
$\Rightarrow | z_{1} - z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} - 2 R e (| z_{1} | | z_{2} | e^{i (arg z_{1} - arg z_{2})}) [∵ | ¯ ¯ ¯ z | = | z |$
$| z_{1} - z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} - 2 | z_{1} | | z_{2} | cos (θ_{1} - θ_{2})$
where $θ_{1} = arg (z_{1})$ and $θ_{2} = arg (z_{2})$ . $[∵ θ_{1} - θ_{2} = 0]$
$\Rightarrow | z_{1} - z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} - 2 | z_{1} | | z_{2} |$
$= (| z_{1} | - | z_{2} |)^{2}$
$\Rightarrow | z_{1} - z_{2} | = | | z_{1} | - | z_{2} | |$

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If for a complex number z1 and z2, arg(z1)−arg(z2)=0, then |z1−z2| is equal to

If for a complex number $z_{1}$ and $z_{2}$ , $arg (z_{1}) - arg (z_{2}) = 0$ , then $| z_{1} - z_{2} |$ is equal to