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Solution


Q.

Let z∈C be such that z2+3iz–2+i=2+3i. Then the sum of all possible values of z2 is          [2025]
1 –19 + 2i  
2 19 + 2i  
3 19 – 2i  
4 –19 –2i  

Ans.

(4)

We have, z2+3iz–2+i=2+3i

⇒ z2+3i=z(2+3i)+(2+3i)(–2+i)

⇒ z2+3i=z(2+3i)–7–4i

⇒ z2–z(2+3i)+7+7i=0

It is a quadratic equation in z

∴   Sum of roots = z1+z2=2+3i

Product of roots = z1z2=7+7i

Now, z12+z22=(z1+z2)2–2z1z2

           = (2+3i)2–2(7+7i)

           = 4 – 9 + 12i – 14 –14i

           = –19 – 2i.