If Z1,Z2,Z3 be three unimodular complex numbers such thatZ12Z2Z3+Z22Z1Z3+Z32Z1Z2+1=0then Z1+Z2+Z3 can take the value equal to.

Here, z1=z2=z3=1but, |z|=1⇒zz¯=1Also given z13+z23+z33+z1z2z3=0⇒z13+z23+z33−3z1z2z3=−4z1z2z3⇒z1z2z3z12+z22+z32−∑z1z2=−4z1z2z3Let z1+z2+z3=z⇒z¯+z¯2+z¯3=z¯=1z¯1+1z¯2+1z¯3∴Σz1Σz12−Σz1z2=−4z1z2z3⇒Σz1Σz12−3Σz1z2=−4z1z2z3⇒zz2−3Σz1z2=−4z1z2z3⇒z3=z1z2z33zΣ1z1−4⋅⇒z3=z1z2z3(3zz¯−4)⇒|z|3=|3|z2−4∣ if |z|≥23, then |z|3=3|z|2−4⇒|z|3−3|z|2+4=0⇒(|z|−2)2(|z|+1)=0⇒|z|=2⇒z1+z2+z3=2If 0≤|z|≤23,then |z|3=4−3|z|2⇒|z|3+3|z|2−4=0⇒(|z|−1)(|z|+2)2=0⇒|z|=1⇒z1+z2+z3=1

If Z1,Z2,Z3 be three unimodular complex numbers such thatZ12Z2Z3+Z22Z1Z3+Z32Z1Z2+1=0then Z1+Z2+Z3 can take the value equal to.

Here, z1=z2=z3=1but, |z|=1⇒zz¯=1Also given z13+z23+z33+z1z2z3=0⇒z13+z23+z33−3z1z2z3=−4z1z2z3⇒z1z2z3z12+z22+z32−∑z1z2=−4z1z2z3Let z1+z2+z3=z⇒z¯+z¯2+z¯3=z¯=1z¯1+1z¯2+1z¯3∴Σz1Σz12−Σz1z2=−4z1z2z3⇒Σz1Σz12−3Σz1z2=−4z1z2z3⇒zz2−3Σz1z2=−4z1z2z3⇒z3=z1z2z33zΣ1z1−4⋅⇒z3=z1z2z3(3zz¯−4)⇒|z|3=|3|z2−4∣ if |z|≥23, then |z|3=3|z|2−4⇒|z|3−3|z|2+4=0⇒(|z|−2)2(|z|+1)=0⇒|z|=2⇒z1+z2+z3=2If 0≤|z|≤23,then |z|3=4−3|z|2⇒|z|3+3|z|2−4=0⇒(|z|−1)(|z|+2)2=0⇒|z|=1⇒z1+z2+z3=1

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If Z₁,Z₂,Z₃be three unimodular complex numbers such that $\frac{Z_{1}^{2}}{Z_{2} Z_{3}} + \frac{Z_{2}^{2}}{Z_{1} Z_{3}} + \frac{Z_{3}^{2}}{Z_{1} Z_{2}} + 1 = 0$ then $|Z_{1} + Z_{2} + Z_{3}|$ can take the value equal to.

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Detailed Solution

Here, $|z_{1}| = |z_{2}| = |z_{3}| = 1$ but, $| z | = 1 \Rightarrow z \bar{z} = 1$
Also given $z_{1}^{3} + z_{2}^{3} + z_{3}^{3} + z_{1} z_{2} z_{3} = 0$
$\begin{array}{l} \Rightarrow z_{1}^{3} + z_{2}^{3} + z_{3}^{3} - 3 z_{1} z_{2} z_{3} = - 4 z_{1} z_{2} z_{3} \\ \Rightarrow (z_{1} z_{2} z_{3}) (z_{1}^{2} + z_{2}^{2} + z_{3}^{2} - \sum z_{1} z_{2}) = - 4 z_{1} z_{2} z_{3} \end{array}$
Let $z_{1} + z_{2} + z_{3} = z \Rightarrow \bar{z} + {\bar{z}}_{2} + {\bar{z}}_{3} = \bar{z} = \frac{1}{{\bar{z}}_{1}} + \frac{1}{{\bar{z}}_{2}} + \frac{1}{{\bar{z}}_{3}}$
$\begin{array}{l} ∴ {Σz}_{1} ({Σz}_{1}^{2} - {Σz}_{1} z_{2}) = - 4 z_{1} z_{2} z_{3} \\ \Rightarrow {Σz}_{1} ({({Σz}_{1})}^{2} - 3 {Σz}_{1} z_{2}) = - 4 z_{1} z_{2} z_{3} \\ \Rightarrow z (z^{2} - 3 {Σz}_{1} z_{2}) = - 4 z_{1} z_{2} z_{3} \\ \Rightarrow z^{3} = z_{1} z_{2} z_{3} (3 zΣ \frac{1}{z_{1}} - 4) \cdot \Rightarrow z^{3} = z_{1} z_{2} z_{3} (3 z \bar{z} - 4) \\ \Rightarrow | z |^{3} = {| 3 | z|}^{2} - 4 ∣ if | z | \geq \frac{2}{\sqrt{3}}, then | z |^{3} = 3 | z |^{2} - 4 \\ \Rightarrow | z |^{3} - 3 | z |^{2} + 4 = 0 \Rightarrow (| z | - 2)^{2} (| z | + 1) = 0 \\ \Rightarrow | z | = 2 \Rightarrow |z_{1} + z_{2} + z_{3}| = 2 \end{array}$
If $0 \leq | z | \leq \frac{2}{\sqrt{3}}$ ,then $| z |^{3} = 4 - 3 | z |^{2}$ $\begin{array}{l} \Rightarrow | z |^{3} + 3 | z |^{2} - 4 = 0 \Rightarrow (| z | - 1) (| z | + 2)^{2} = 0 \\ \Rightarrow | z | = 1 \Rightarrow |z_{1} + z_{2} + z_{3}| = 1 \end{array}$