If z_1, z_2, . . z_n are complex numbers such that |z_1|=|z_2|= . .=|z_n|=1 , then |z_1+z_2+ . .+z_n| is equal to

Correct answer: C. |(1)/(z_1)+(1)/(z_2)+ . .+(1)/(z_n)| |z_1|=|z_2|=|z_3|= .=|z_n|=1 z_1 z _1=z_2 z _2=z_3 z _3= =z_n z _n=1 z _1= 1 z , z _2= 1 z_2 , z _3= 1 z_3 , , z _ n = 1 z_ n Now, |z_1+z_2+z_3+ .+z_n|=| z _1+ z _2+ + z _n|| z _1+ z _2+ z _3+ .+ z _n|=|(1)/(z_1)+(1)/(z_2)+(1)/(z_3)+ +(1)/(z_n)|

BITSAT 2024Paper 2

Exam: BITSAT
Year: 2024
Paper: Paper 2
Subject: Mathematics
Chapter: Complex Number

BITSAT 2024 · Paper 2 · Mathematics · Complex Number

If z_1, z_2, . . z_n are complex numbers such that |z_1|=|z_2|= ....

z_1, z_2, \ldots . . z_n

, then \left|z_1+z_2+\ldots . .+z_n\right|

is equal to

Options

A $\left|z_1 z_2 z_3 \ldots . . z_n\right|$
B $\left|z_1\right|+\left|z_2\right|+\ldots . .+\left|z_n\right|$
C $\left|\frac{1}{z_1}+\frac{1}{z_2}+\ldots . .+\frac{1}{z_n}\right|$
Dn

Correct answer

C. $\left|\frac{1}{z_1}+\frac{1}{z_2}+\ldots . .+\frac{1}{z_n}\right|$

Step-by-step solution

\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=\ldots \ldots .=\left|z_n\right|=1\Rightarrow z_1 \bar{z}_1=z_2 \bar{z}_2=z_3 \bar{z}_3=\ldots \ldots=z_n \bar{z}_n=1\overline{\mathrm{z}}_1=\frac{1}{\mathrm{z}}, \overline{\mathrm{z}}_2=\frac{1}{\mathrm{z}_2}, \overline{\mathrm{z}}_3=\frac{1}{\mathrm{z}_3}, \ldots \ldots, \overline{\mathrm{z}}_{\mathrm{n}}=\frac{1}{\mathrm{z}_{\mathrm{n}}}

Now,

\left|z_1+z_2+z_3+\ldots \ldots .+z_n\right|=\left|\bar{z}_1+\bar{z}_2+\ldots \ldots+\bar{z}_n\right|\left|\bar{z}_1+\bar{z}_2+\bar{z}_3+\ldots .+\bar{z}_n\right|=\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+\ldots \ldots+\frac{1}{z_n}\right|

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