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If z_1=5+12i and |z_2|=4, then

Question

If $z_{1} = 5 + 12 i$ and $∣ z_{2} ∣ = 4$ , then

Maximum

(∣ z_{1} + i z_{2} ∣) = 17

Minimum

(∣ z_{1} + (1 + i) z_{2} ∣) = 13 - 42

Minimum

∣ ∣ \frac{z _{1}}{z _{2} + \frac{4}{z _{2}}} ∣ ∣ = \frac{13}{4}

Maximum

∣ ∣ \frac{z _{1}}{z _{2} + \frac{4}{z _{2}}} ∣ ∣ = \frac{13}{3}

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Elizabeth Discussed

z_{1} = 5 + 12 i

and

∣ z_{2} ∣ = 4

, then

5 mins ago

Discuss this question LIVE

5 mins ago

Text solutionVerified

z_{1} = 5 + 12 i, ∣ z_{2} ∣ = 4

∣ z_{1} + i z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣ = 13 + 4 = 17

∴ ∣ z_{1} + (1 + i) z_{2} ∣ \geq ∣ ∣ z_{1} ∣ - ∣1 + i ∣∣ z_{2} ∣ ∣

= 13 - 42

∴ min (∣ z_{1} + (1 + i) z_{2} ∣) = 13 - 42

∣ ∣ z_{1} + \frac{4}{z _{2}} ∣ ∣ \leq ∣ z_{2} ∣ + \frac{4}{∣ z _{2} ∣} = 4 + 1 = 5

∣ ∣ z_{2} + \frac{4}{z _{2}} ∣ ∣ \geq ∣ z_{2} ∣ - \frac{4}{∣ z _{2} ∣} = 4 - 1 = 3

∴

max

∣ ∣ \frac{z _{1}}{z _{2} + \frac{4}{z _{2}}} ∣ ∣ = \frac{13}{3}

and min

∣ ∣ \frac{z _{1}}{z _{2}} + \frac{4}{z _{2}} ∣ ∣ = \frac{13}{5}

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