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Properties of nth Root of a Complex Number

Let z1,z2,z3,...

Question

Let $z_{1}, z_{2}, z_{3}, z_{4}, z_{5}$ and $z_{6}$ be complex numbers lying on a unit circle with centre $(0, 0) .$ If $ω = (6 \sum k = 1 z_{k}) (6 \sum k = 1 \frac{1}{z_{k}}),$ then

ω

is purely imaginary.

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ω

is purely real.

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Maximum possible value of

| ω |

36.

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Maximum possible value of

| ω |

6.

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Solution

The correct option is C Maximum possible value of $| ω |$ is $36.$
$| z_{k} | = 1 \Rightarrow \frac{1}{z_{k}} =_{k}$
$ω = (6 \sum k = 1 z_{k}) (6 \sum k = 1 \frac{1}{z_{k}}) ω = (z_{1} + z_{2} + \dots + z_{6}) (\frac{1}{z_{1}} + \frac{1}{z_{2}} + \dots + \frac{1}{z_{6}}) = (z_{1} + z_{2} + \dots + z_{6}) (_{1} +_{2} + \dots +_{6}) = (z_{1} + z_{2} + \dots + z_{6}) (¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z_{1} + z_{2} + \dots + z_{6})$
$ω = {| z_{1} + z_{2} + \dots + z_{6} |}_{1}^{2}$ which is purely real.

$0 \leq | z_{1} + z_{2} + \dots + z_{6} | \leq | z_{1} | + | z_{2} | + \dots + | z_{6} | \Rightarrow 0 \leq (| z_{1} + z_{2} + \dots + z_{6} |)^{2} \leq (| z_{1} | + | z_{2} | + \dots + | z_{6} |)^{2} \Rightarrow 0 \leq | ω | \leq 36$

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Similar questions

z_{1}, z_{2}, z_{3}, z_{4}, z_{5}

z_{6}

(0, 0) .

ω = (6 \sum k = 1 z_{k}) (6 \sum k = 1 \frac{1}{z_{k}}),

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

z_{i} + z_{j} z_{k},

i, j, k \in {1, 2, 3}

i \neq j \neq k

4 (z_{1} \times z_{2} \times z_{3})

If $z_{1}, z_{2}$ are conjugate complex numbers, and $z_{3}, z_{4}$ are also conjugate, then show that

$arg \frac{z_{3}}{z_{2}} + arg \frac{z_{1}}{z_{4}} = 0$ .

If $(z_{1}, z_{2})$ and $(z_{3}, z_{4})$ are two pairs of conjugate complex numbers, then show that

$arg \frac{z_{1}}{z_{3}} + arg \frac{z_{2}}{z_{4}} = 0$ .

z_{1}, z_{2}, z_{3}, z_{4}, z_{5}

z^{5} + z^{4} + z^{3} + z^{2} + z + 1 = 0

∣ ∣ ∣ ∣ 5 \sum i = 1 z_{i}^{4} ∣ ∣ ∣ ∣

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