Prove that - 1- z 1 z- 2 z 1 - z 2 - -1- z 1 - -1- z 2 -

Proceeding by squaring ( 1 | z _ 1 | ^ 2 ) ( 1 | z _ 2 | ^ 2 ) lt;0 i.e., ive .Above will be true if | z _ 1 | lt;1 lt;| z _ 2 | . ...

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Inequality of 2 Complex Numbers

Prove that ...

Question

Prove that $∣ ∣ ∣ \frac{1 - z_{1} {¯ z}_{2}}{z_{1} - z_{2}} ∣ ∣ ∣ < 1 | z_{1} | < 1 < | z_{2} |$

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| z_{1} | + | z_{2} | = ∣ ∣ ∣ \frac{1}{2} (z_{1} + z_{2}) + \sqrt{z_{1} z_{2}} ∣ ∣ ∣ + ∣ ∣ ∣ \frac{1}{2} (z_{1} + z_{2}) - \sqrt{z_{1} z_{2}} ∣ ∣ ∣

∣ ∣ ∣ \frac{z_{1} - z_{2}}{1 -_{1} z_{2}} ∣ ∣ ∣ < 1

| z_{1} | < 1

| z_{2} | < 1

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

[z_{1} + z_{2} + . . . . + z_{n}] = ∣ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + . . . . + \frac{1}{z_{n}} ∣ ∣ ∣

z_{1}, z_{2}, . . . . ., z_{n}

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

| ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z_{1} + z_{2} + . . . . . + z_{n} | = ∣ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + . . . . . . + \frac{1}{z_{n}} ∣ ∣ ∣

z_{1}, z_{2}

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

| z_{1} + 1 | + | z_{2} + 1 | + | z_{1} z_{2} + 1 | \geq 2

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