If - z 1-2- - z 2 be the point in which sartisfies the condition z 2- z 2- i z 2- z 2-8 -2- then the minimum value of - z 1- z 2- is

|z_1| = 2 ⇒ x^2+y^2 = 4 (z_2+z_2) i(z_2 z_2) = 8√(2) ⇒ x+y = 4√(2) nbsp;AB = OB r AB = 2 ...

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Equation of Conics in Complex Form

If | z 1|=2, ...

Question

If $| z_{1} | = 2, & z_{2}$ be the point in which sartisfies the condition $(z_{2} +_{2}) - i (z_{2} -_{2}) = 8 \sqrt{2},$ then the minimum value of $| z_{1} - z_{2} |$ is

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z_{1} = 5 + 12 i

| z_{2} | = 4

If arg ( $Z_{1}$ ) = $α$ , arg ( $Z_{2}$ ) = $β$ , the value of $| Z_{1} + Z_{2} |^{2}$ is

z_{1}

z_{2}

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

z_{1} & z_{2}

a r g (\frac{z_{1} + z_{2}}{z_{1} - z_{2}}) = \frac{π}{2}

| z_{1} + z_{2} | \neq | z_{1} - z_{2} |

z_{1}, z_{2}, z_{1} + z_{2}

[\frac{z_{1}}{z_{2}}]

=

\frac{π}{2},

∣ ∣ \frac{z_{1} + z_{2}}{z_{1} - z_{2}} ∣ ∣ .

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