Question

What is the number of distinct solutions of the equation \[{z^2} + \left| z \right| = 0\] (where \[z\] is a complex number)
A.One
B.Two
C.Three
D.Five

Answer 1

Hint: First we will assume the real part and imaginary part of the complex to be any variable. Then we will find the square of that complex number. Then we will substitute all the values in the given equation and equate the imaginary and real part of the simplified complex number with zero. We will find the value of the variables from the obtained equations.

Complete step-by-step answer:
Let the complex number \[z\] be \[x + iy\] i.e.
\[z = x + iy\] ……. \[\left( 1 \right)\]
Now, we will find the square of the assumed complex number.
On squaring both sides, we get
\[ \Rightarrow {\left( z \right)^2} = {\left( {x + iy} \right)^2}\]
Using the algebraic identity, \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], we get
\[ \Rightarrow {z^2} = {\left( x \right)^2} + {\left( {iy} \right)^2} + 2 \times x \times iy\]
Applying exponents on the bases, we get
\[ \Rightarrow {z^2} = {x^2} - {y^2} + 2ixy\]
Now, we will find the modulus of the complex number.
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
Now, we will substitute all the values in the given equation, \[{z^2} + \left| z \right| = 0\].
\[ \Rightarrow {x^2} - {y^2} + 2ixy + \sqrt {{x^2} + {y^2}} = 0\]
Now, we will separate the imaginary and real parts of the complex numbers here.
\[ \Rightarrow {x^2} - {y^2} + \sqrt {{x^2} + {y^2}} + i\left( {2xy} \right) = 0\]
We can write the right-hand side of the equation as
\[ \Rightarrow {x^2} - {y^2} + \sqrt {{x^2} + {y^2}} + i\left( {2xy} \right) = 0 + 0i\]
Now, we will individually equate real and imaginary parts with zero. Therefore we get
\[ \Rightarrow {x^2} - {y^2} + \sqrt {{x^2} + {y^2}} = 0\] ……….. \[\left( 2 \right)\]
And
 \[ \Rightarrow \left( {2xy} \right) = 0\] …….. \[\left( 3 \right)\]
First we will solve the equation \[\left( 3 \right)\].
\[ \Rightarrow \left( {2xy} \right) = 0\]
This is possible when \[x = 0\] or \[y = 0\].
Now, we will put \[x = 0\] in equation \[\left( 2 \right)\].
\[ \Rightarrow {0^2} - {y^2} + \sqrt {{0^2} + {y^2}} = 0\]
On further simplification, we get
\[ \Rightarrow - {y^2} + \left| y \right| = 0\]
On simplifying the modulus, we get
\[ \Rightarrow - {y^2} + y = 0\]
And
\[ \Rightarrow - {y^2} - y = 0\]
Now, first we will solve \[ - {y^2} + y = 0\]
\[ \Rightarrow y\left( { - y + 1} \right) = 0\]
Therefore, we get
\[ \Rightarrow \left( { - y + 1} \right) = 0\] and \[y = 0\]
On further simplification, we get
\[ \Rightarrow y = 1\] and \[y = 0\]
Now, we will solve \[ - {y^2} - y = 0\]
On factoring the equation, we get
\[ \Rightarrow y\left( {y + 1} \right) = 0\]
On further simplification, we get
\[ \Rightarrow y = 0\] and \[y + 1 = 0\]
Therefore, we get
\[ \Rightarrow y = 0\] and \[y = - 1\]
Now, we will put \[y = 0\] in equation \[\left( 2 \right)\].
\[ \Rightarrow {x^2} - {0^2} + \sqrt {{x^2} + {0^2}} = 0\]
On further simplification, we get
\[ \Rightarrow {x^2} + \left| x \right| = 0\]
This is possible only when \[x = 0\] as both the terms are positive.
Therefore, the values that we have got are:-
When \[x = 0\], then \[y = 0\] , \[y = 1\] and \[y = - 1\].
When \[y = 0\] , then \[x = 0\]
Thus, we have got the solutions by putting the values in the complex number as \[0 + 0i\], \[0 + i\] and \[0 - i\].
Hence, there are a total three distinct solutions of the given equation.
Hence, the correct option is option C.

Note: Here we have got the value of \[x\] as 0 from the equation \[{x^2} + \left| x \right| = 0\] because both the terms are positive here. First-term is the square of \[x\] and the second term is the modulus of \[x\]. We know that the square of any number gives a positive number and = modulus of any number always gives a positive number. Thus, the sum of two positive terms is zero only when both the terms are equal to zero.