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Substitution Method to Remove Indeterminate Form

If |z1+z2|=...

Question

If $| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$ , then $a r g z_{1} - a r g z_{2} =$

A

\frac{π}{2}

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B

\frac{- π}{2}

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C

0

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D

- π

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Solution

The correct option is B $0$
$| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$
On squaring,
$| z_{1} |^{2} + | z_{2} |^{2} + 2 | z_{1} | | z_{2} | c o s θ = | z_{1} |^{2} + | z_{2} |^{2} + 2 | z_{1} | | z_{2} |$
where, $θ = a r g z_{1} - a r g z_{2}$
$\Rightarrow cos θ = 1$
$\Rightarrow θ = 0$
Hence, option C is correct.

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Similar questions

| z_{1} - z_{2} | = | z_{1} | + | z_{2} |

, then prove that

a r g (z_{1}) - a r g (z_{2}) = π

z_{1}, z_{2}

are complex numbers then the correct match List- I to List II is:

List - I	List - II
A) arg z_1z_2	1) $a r g z_{1} - a r g z_{2}$
B) $a r g z_{1}_{2}$	2) $a r g z_{1} - a r g z_{2} = \frac{π}{2}$
C) $\| z_{1} + z_{2} \| = \| z_{1} - z_{2} \|$	3) $a r g z_{1} = a r g z_{2}$
D) $\| z_{1} + z_{2} \|^{2}$	4) $a r g z_{1} + a r g z_{2}$
E) $\| z_{1} + z_{2} \| = \| z_{1} \| + \| z_{2} \|$	5) $r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s (θ_{1} - θ_{2})$

a r g (\frac{z_{1}}{z_{2}}) = \frac{π}{2}

, then find the value of

∣ ∣ ∣ \frac{z_{1} + z_{2}}{z_{1} - z_{2}} ∣ ∣ ∣

.

z_{1}

z_{2}

are two complex numbers such that

| z_{1} | = | z_{2} |

a r g z_{1} + a r g z_{2} = π

z_{1}

z_{2}

| z_{1} + z_{2} | = | z_{1} - z_{2} |

a r g z_{1} - a r g z_{2} =

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