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Estimating the Square Root

If z1 = 2 -...

Question

If $z_{1} = 2 - 5 i, z_{2} = 3 - i$ , then magnitude of projection of $z_{1}$ on $z_{2}$ is .............

\frac{11}{\sqrt{10}}

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\frac{11}{\sqrt{5}}

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\frac{7}{\sqrt{10}}

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\frac{9}{\sqrt{5}}

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Solution

The correct option is A $\frac{11}{\sqrt{10}}$

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Similar questions

z_{1} = \frac{2 \sqrt{3} + i 6 \sqrt{7}}{6 \sqrt{7} + i 2 \sqrt{3}}

z_{2} = \frac{\sqrt{11} + i 3 \sqrt{13}}{3 \sqrt{13} + i \sqrt{11}}

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

z_{1}, z_{2}

| z_{1} | = \sqrt{2}, | z_{2} | = \sqrt{3} and | z_{1} + z_{2} | = \sqrt{5 - 2 \sqrt{3}}

| arg (z_{1}) - arg (z_{2}) |

z_{1} = 7 + 24 i

| z_{2} | = 5,

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

3

\frac{3}{| z_{2} - z_{3} |} = \frac{4}{| z_{3} - z_{1} |} = \frac{5}{| z_{1} - z_{2} |}

\frac{9}{z_{2} - z_{3}} + \frac{16}{z_{3} - z_{1}} + \frac{25}{z_{1} - z_{2}}

z_{1}

z_{2}

z_{3}

\frac{1}{∣ z_{1} - z_{2} ∣} = \frac{3}{∣ z_{2} - z_{3} ∣} = \frac{5}{∣ z_{1} - z_{3} ∣}

\frac{1}{z_{1} - z_{2}} + \frac{9}{z_{2} - z_{3}} + \frac{25}{z_{3} - z_{1}}

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