Question

If z is a complex number then \[{z^2} + {\bar z^2} = 2\] represents a
A) A circle
B) A Straight line
C) A hyperbola
D) An ellipse

Answer 1

Hint: If z is a complex number then \[\bar z\] is represented as its conjugate. Take any general representation of a complex number and find its conjugate then put it in the equation given to solve it to the end. You will get a final equation.

Complete Step by Step Solution:
Let us represent our complex number z as \[x + iy\] clearly the conjugate \[\bar z\] will be \[x - iy\]. Now for solving this put all the values in the equation given. Thus, the equation becomes,
\[\begin{array}{l}
{z^2} + {{\bar z}^2} = 2\\
 \Rightarrow {(x + iy)^2} + {(x - iy)^2} = 2\\
 \Rightarrow {x^2} + 2ixy + {(iy)^2} + {x^2} - 2ixy + {(iy)^2} = 2\\
 \Rightarrow 2{x^2} + 2{i^2}{y^2} = 2\\
Now,i = \sqrt { - 1} \\
{i^2} = - 1\\
\therefore 2{x^2} + 2 \times ( - 1){y^2} = 2\\
 \Rightarrow {x^2} - {y^2} = 1
\end{array}\]
Now the general equation of an hyperbola is given by \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] clearly a and b are 1 here respectively
Therefore option C is correct.

Note: We have used \[{(a + b)^2} = {a^2} + 2ab + {b^2}\] to solve \[{(x + iy)^2}\& {(x - iy)^2}\] . You can also do this question by taking \[z = r(\cos \theta + i\sin \theta )\] So as you are assuming the complex number in polar form. So you will also get the final equation in polar form so remember to convert it according to your convenience.