# Q. The number of solutions of the equation `z^2` + `|z|^2` = `0` where `z` is a complex number:- A) one B) two C) three D) infinitely many

degeneratecircle | High School Teacher | (Level 2) Associate Educator

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Alternate Solution:

Use the facts that `|z|^2=zbarz` and `z+barz=2Re(z),` where `Re(z)` denotes the real part of `z` (i.e., if `z=a+bi,` then `Re(z)=a`).

Then we have

`0=z^2+|z|^2=z^2+zbarz=z(z+barz)=2zRe(z),`

so either `z=0` or ` ` `Re(z)=0.` Thus the (infinite) solution set is

`{bi | bin RR},` in other words precisely those complex numbers that lie on the imaginary axis.

Sources:

justaguide | College Teacher | (Level 2) Distinguished Educator

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Let the complex number z = a + i*b

`|z| = sqrt(a^2 + b^2)`

`z^2 + |z|^2 = 0`

=> `(a + i*b)^2 + (sqrt(a^2 +b^2))^2 = 0`

=> `a^2 - b^2 + 2*i*a*b + a^2 + b^2 = 0`

=> `2a^2 + 2*i*a*b = 0`

=> `a(a + i*b) = 0`

a = 0, b can take on any value

a + i*b = 0

=> a = 0, b = 0

The given equation can have an infinite number of solutions.

The correct answer is option D