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