मराठी

If |z1| = |z2| = ... = |zn| = 1, then show that |z1 + z2 + z3 + ... + zn| = |1z1+1z2+1z3+...+1zn|. - Mathematics

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प्रश्न

If |z1| = |z2| = ... = |zn| = 1, then show that |z1 + z2 + z3 + ... + zn| = `|1/z_1 + 1/z_2 + 1/z_3 + ... + 1/z_n|`.

बेरीज
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उत्तर

We have |z1| = |z2| = ... = |zn| = 1

⇒ |z1|2 = |z2|2 = ... = |zn|2 = 1  ......(i)

⇒ `z_1 barz_1 = z_2 barz_2 = ... = z_n barz_n` = 1  .....`[because zbarz = |z|^2]`

⇒ z1 = `1/barz_1, z_2 = 1/barz_2 = ... = z_n = 1/barz_n`

L.H.S. |z1 + z2 + z3 + ... + zn|

= `|(z_1barz_1)/barz_1 + (z_2barz_2)/barz_2 + (z_3barz_3)/barz_3 + ... + (z_nbarz_n)/barz_n|`

= `||z_1|^2/barz_1 + (|z_2|^2)/barz_2 + (|z_3|^2)/barz_3 + ... + (|z_n|^2)/barz_n|`  ......`[zbarz = |z|^2]`

= `|1/barz_1 + 1/barz_2 + 1/barz_3 + ... + 1/barz_n|`  ......[Using (i)]

= `|bar(1/z_1 + 1/z_2 + 1/z_3 + ... + 1/z_n)|`  .....`[because barz_1 + barz_2 = bar(z_1 + z_2)]`

= `|1/z_1 + 1/z_2 + 1/z_3 + ... + 1/z_n|`  ....`[because |z| = |barz|]`

L.H.S. = R.H.S.

Hence proved.

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पाठ 5: Complex Numbers and Quadratic Equations - Exercise [पृष्ठ ९२]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 5 Complex Numbers and Quadratic Equations
Exercise | Q 19 | पृष्ठ ९२

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