If -z1-z2-z3-zn-1 then prove that-1-z1-1-22-1-z3-1-zn-z1-z2-z3-zn-

We have |z_1|=|z_2|=|z_3|=...=|z_n|=1 ⇒ |z_1|^2=|z_2|^2=|z_3|^2= ... = |z_n|^2=1 ⇒ z_1z̅_̅1̅=1, z_2z̅_̅2̅=1, z_3z̅_̅3̅=1, ..., z_nz̅_̅n̅=1 ⇒1/z_1=z̅_̅1̅, 1 ...

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Byju's Answer

Standard XII

Mathematics

Definition of Relations

If |z1|=|z2|=...

Question

If $| z_{1} | = | z_{2} | = | z_{3} | = . . . = | z_{n} | = 1$ then prove that
$∣ ∣ ∣ \frac{1}{z_{1} + \frac{1}{z_{2}} + \frac{1}{z_{3}}} + . . . + \frac{1}{z_{n}} ∣ ∣ ∣ = | z_{1} + z_{2} + z_{3} + . . . + z_{n} | .$

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Solution

We have

$| z_{1} | = | z_{2} | = | z_{3} | = . . . = | z_{n} | = 1$

$\Rightarrow | z_{1} |^{2} = | z_{2} |^{2} = | z_{3} |^{2} = . . . = | z_{n} |^{2} = 1$

$\Rightarrow z_{1}_{1} = 1, z_{2}_{2} = 1, z_{3}_{3} = 1, . . ., z_{n}_{n} = 1$

$\Rightarrow \frac{1}{z_{1}} =_{1}, \frac{1}{z_{2}} =_{2}, \frac{1}{z_{3}} =_{3}, . . ., \frac{1}{z_{n}} =_{n}$

$\Rightarrow ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} + . . . + \frac{1}{z_{n}} ∣ ∣ = |_{1} +_{2} +_{3} + . . . +_{n} |$

$= ¯ | z_{1} + z_{2} + z_{3} + . . . + z_{n} |$

$= | z_{1} + z_{2} + z_{3} + . . . + z_{n} |$ $[∵ | ¯ z | = | z |]$

$∴ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} + . . . + \frac{1}{z_{n}} ∣ ∣ = | z_{1} + z_{2} + z_{3} + . . . + z_{n} |$

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Q. If

| z_{1} | = | z_{2} | = | z_{3} | = . . . . . . . = | z_{n} | = 1,

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| z_{1} + z_{2} + z_{3} + . . . . . . . + z_{n} | =

Q. If

| z_{1} | = | z_{2} | = . . . = | z_{n} | = 1

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Q. If

z_{1}, z_{2}, z_{3}

are complex numbers such that

| z_{1} | = | z_{2} | = | z_{3} | = ∣ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ ∣ = 1

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are complex number such that

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

| z_{3} |

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If |z1|=|z2|=|z3|=...=|zn|=1 then prove that ∣∣∣1z1+1z2+1z3+...+1zn∣∣∣=|z1+z2+z3+...+zn|.

If $| z_{1} | = | z_{2} | = | z_{3} | = . . . = | z_{n} | = 1$ then prove that
$∣ ∣ ∣ \frac{1}{z_{1} + \frac{1}{z_{2}} + \frac{1}{z_{3}}} + . . . + \frac{1}{z_{n}} ∣ ∣ ∣ = | z_{1} + z_{2} + z_{3} + . . . + z_{n} | .$