If $| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$ where $z_{1}$ and $z_{2}$ are different non - zero complex number, then ?

R e (\frac{z_{1}}{z_{2}}) = 0

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I m (\frac{z_{1}}{z_{2}}) = 0

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z_{1} + z_{2} = 0

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None

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Solution

The correct option is C $R e (\frac{z_{1}}{z_{2}}) = 0$
$l e t z_{1} = a + i b a n d z_{2} = c + i d \Rightarrow | z_{1} + z_{2} | = | z_{1} | + | z_{2} | \Rightarrow | a + i b + c + i d | = | a + i b | + | c + i d | \Rightarrow | a + c + i (b + d) | = | a + i b | + | c + i d | \Rightarrow \sqrt{{(a + c)}^{2} + {(b + d)}^{2}} = \sqrt{a^{2} + b^{2}} + \sqrt{c^{2} + d^{2}} s q u a r i n g b o t h s i d e s \Rightarrow {(a + c)}^{2} + {(b + d)}^{2} = a^{2} + b^{2} + c^{2} + d^{2} + 2 \sqrt{a^{2} + b^{2}} . \sqrt{c^{2} + d^{2}} \Rightarrow 2 a c + 2 b d = 2 \sqrt{a^{2} + b^{2}} . \sqrt{c^{2} + d^{2}} \Rightarrow a c + b d = \sqrt{a^{2} + b^{2}} . \sqrt{c^{2} + d^{2}} a g a i n s q u a r i n g b o t h s i d e s \Rightarrow {(a c)}^{2} + {(b d)}^{2} + 2 a b c d = (a^{2} + b^{2}) (c^{2} + d^{2}) \Rightarrow {(a c)}^{2} + {(b d)}^{2} + 2 a b c d = {(a c)}^{2} + {(a d)}^{2} + {(b c)}^{2} + {(b d)}^{2} \Rightarrow {(a d)}^{2} + {(b c)}^{2} - 2 a b c d = 0 \Rightarrow {(a d - b c)}^{2} = 0 \Rightarrow a d - b c = 0 a d = b c ⟶ (1) N o w \frac{z_{1}}{z_{2}} = \frac{a + i b}{c + i d} = \frac{(a + i b) (c - i d)}{(c + i d) (c - i d)} = \frac{a c - i a d + i b c - i^{2} b d}{c^{2} + d^{2}} = \frac{a c + b d + i (b c - a d)}{c^{2} + d^{2}} = \frac{a c + b d}{c^{2} + d^{2}} + 0 u s i n g e q u a t i o n (1) I m a g i n a r y p a r t i s z e r o a s t h e r e i s o n l y r e a l p a r t ∴ I m (\frac{z_{1}}{z_{2}}) = 0$

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Similar questions

If $| z_{1} + z_{2} |^{2}$ = $| z_{1} - z_{2} |^{2}$ where $z_{1}$ and $z_{2}$ are non-zero complex numbers, then :

If $| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$ where $z_{1}$ and $z_{2}$ are different non-zero complex numbers, then :

z_{1}

z_{2}

z_{1} \neq z_{2}

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

R e (z_{1}) > 0

l m (z_{2}) < 0

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

z_{1}, z_{2}

| z_{1} + z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} .

z_{1}

z_{2}

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

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