Question

For real parameter t, the locus of the complex number $z=\left( 1-{{t}^{2}} \right)+i\sqrt{1+{{t}^{2}}}$ in the complex plane is
(a) an ellipse
(b) a parabola
(c) a circle
(d) a hyperbola

Answer 1

Hint: To find the locus in such types of questions, we assume the complex number z as x + iy. Then by comparing the real and imaginary part on both the sides of the obtained equation, we find the value of x and y in terms of the parameter t. Then we try to eliminate it using both the equations. The equation we obtain after eliminating the parameter t is the equation of the locus of the complex number. Using this information, we can solve this question.

Complete step-by-step answer:
In the question, we are given a complex number $z=\left( 1-{{t}^{2}} \right)+i\sqrt{1+{{t}^{2}}}$ and we are required to find the locus of this complex number.
Since z is a complex number, let us substitute it as x + iy. So, we get,
$x+iy=\left( 1-{{t}^{2}} \right)+i\sqrt{1+{{t}^{2}}}$
Comparing the real and imaginary part in the above equation, we get,
$x=\left( 1-{{t}^{2}} \right)$ . . . . . . . . . . . . . . (1)
$y=\sqrt{1+{{t}^{2}}}$ . . . . . . . . . . . . . . . (2)
From (1), we can write,
${{t}^{2}}=1-x$
Substituting this value of ${{t}^{2}}$ in equation (2), we get,

& y=\sqrt{1+\left( 1-x \right)} \\\ & \Rightarrow y=\sqrt{2-x} \\\ \end{aligned}$$ Squaring both the sides of the above equation, we get, $$\begin{aligned} & {{y}^{2}}={{\left( \sqrt{2-x} \right)}^{2}} \\\ & \Rightarrow {{y}^{2}}=2-x \\\ \end{aligned}$$ We can observe that the above equation is the equation of a parabola. So, the locus of the complex number z is a parabola. Hence, the answer is option (b). Note: It is a misconception that the locus is an equation. The locus is actually the nature of the curve on which the point (of which we are calculating the locus) is moving. So, when we are required to find the locus, we give our answer as the type of curve and not the equation.