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Exercise 2.9 · Q13

Q.z1,z2z_1, z_2, and z3z_3 are complex numbers such that z1+z2+z3=0z_1+z_2+z_3=0 and ∣z1∣=∣z2∣=∣z3∣=1|z_1|=|z_2|=|z_3|=1 then z12+z22+z32z_1^2+z_2^2+z_3^2 is

(1) 33
(2) 22
(3) 11
(4) 00
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Squaring the given sum z1+z2+z3=0z_1+z_2+z_3=0 relates ∑zk2\sum z_k^2 to ∑zizj\sum z_iz_j; the unimodular condition then shows this pairwise-product sum is also zero, by conjugating the original equation.

Step 1. Square the given relation z1+z2+z3=0z_1+z_2+z_3=0.

(z1+z2+z3)2=0  ⟹  z12+z22+z32+2(z1z2+z2z3+z3z1)=0,(z_1+z_2+z_3)^2=0 \;\Longrightarrow\; z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1)=0,

so

z12+z22+z32=−2(z1z2+z2z3+z3z1).(∗)z_1^2+z_2^2+z_3^2=-2(z_1z_2+z_2z_3+z_3z_1). \quad (\ast)

Step 2. Use ∣zk∣=1|z_k|=1 to write zk‾=1zk\overline{z_k}=\dfrac1{z_k} for each kk.

Step 3. Conjugate the original equation z1+z2+z3=0z_1+z_2+z_3=0.

z1‾+z2‾+z3‾=0  ⟹  1z1+1z2+1z3=0.\overline{z_1}+\overline{z_2}+\overline{z_3}=0 \;\Longrightarrow\; \frac1{z_1}+\frac1{z_2}+\frac1{z_3}=0.

Step 4. Combine over the common denominator z1z2z3z_1z_2z_3.

z2z3+z1z3+z1z2z1z2z3=0  ⟹  z1z2+z2z3+z3z1=0\frac{z_2z_3+z_1z_3+z_1z_2}{z_1z_2z_3}=0 \;\Longrightarrow\; z_1z_2+z_2z_3+z_3z_1=0

(the denominator z1z2z3≠0z_1z_2z_3\ne0 since each ∣zk∣=1≠0|z_k|=1\ne0).

Step 5. Substitute this into (∗)(\ast). …

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