Problem
Let z be a complex number satisfying |z|³ + 2z² + 4\(\boldsymbol{\bar{z}}\) – 8 = 0, where \(\boldsymbol{\bar{z}}\) denotes the complex conjugate of z. Let the imaginary part of z be nonzero.
Match each entry in List-I to the correct entries in List-II.
| List-I | List-II |
| (P) |z|² is equal to | (1) 12 |
| (Q) |z – \(\boldsymbol{\bar{z}}\)|² is equal to | (2) 4 |
| (R) |z|² + |z + \(\boldsymbol{\bar{z}}\)|² is equal to | (3) 8 |
| (S) |z + 1|² is equal to | (4) 10 |
| (5) 7 |
Solution
If we assume z = x + iy, the given equation changes to |x + iy|³ + 2(x + iy)² + 4(x – iy) – 8 = 0.
On separating out the real and imaginary parts, we’ll get (|x + iy|³ + 2x² – 2y² + 4x – 8) + i(4xy – 4y) = 0
This gives us two equations |x + iy|³ + 2x² – 2y² + 4x – 8 = 0 and 4xy – 4y = 0.
The first one is scary – let’s stay away from that for a bit.
The second one is much simpler: 4xy – 4y = 0. The LHS can be factorized, and we get 4y(x – 1) = 0.
Since the imaginary part of z, i.e. y, is non-zero, we’ll get x = 1 as the solution.
Now, let’s subsitute the same in the first equation and see if we get the value of y (and therefore z).
We have, |1 + iy|³ + 2(1)² – 2y² + 4(1) – 8 = 0
This gives |1 + iy|³ = 2(1 + y²). Using the definition of |z|, we’ll get √(1 + y²)³ = 2(1 + y²).
Finally, we’ll get √(1 + y²) = 2, which gives y = ±√3. Therefore, z = 1 ± i√3.
Now, the rest of the problem involves substituting the value of z in each expression and finding its value. Let’s pick each option one by one.
Option P
|z|² equals |1 ± i√3|² or 4. Therefore, (P) matches with (2).
Option Q
|z – \(\boldsymbol{\bar{z}}\)|² equals |±2i√3|² or 12. Therefore, (Q) matches with (1).
Option R
|z|² + |z + \(\boldsymbol{\bar{z}}\)|² equals |1 ± i√3|² + |2|² or 8. Therefore, (R) matches with (3).
Option S
|z + 1|² equals |2 ± i√3|² or 7. Therefore, (S) matches with (5).
Comments
Mixed feelings about this problem. It started off well, but each expression was based on the absolute value. Some variety atleast?
For example, asking the value of |arg(z)| or |arg(z – 1)|. Or, maybe expressions involving z1 and z2, the two values of z that we got from the equation? Yeah, not much effort was put into it. Also, as a result, wasn’t that challenging really. More like a JEE Main problem.
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