Question

Let $z = 1 - t + i\sqrt {{t^2} + t + 2} $, where $t$ is the real parameter. The locus of \[z\] in the argand plane is:
A. A hyperbola
B. An ellipse
C. A straight line
D. None of these

Answer 1

Hint: This is a question of complex analysis. If $z = x + iy$ is the complex number then $x$ is called as the real part of $z$ and $y$ is called as the imaginary part of $z$. Here, $i$ is the imaginary number.
We have to find the equation of the plane, according to the equation we find the argand plane. In the argand plane, $x$-axis is the real axis and $y$ is the imaginary axis.
Compare $z = 1 - t + i\sqrt {{t^2} + t + 2} $ with the complex number$z = x + iy$.
Where $x = 1 - t$ and $y = \sqrt {{t^2} + t + 2} $.
Remove the real parameter $t$ with the help of $x = 1 - t$ and $y = \sqrt {{t^2} + t + 2} $. The equation in terms of $x$and $y$ is the equation of the plane and we can determine the required plane.

Complete step-by-step solution:
Consider the complex number, $z = 1 - t + i\sqrt {{t^2} + t + 2} $, where $t$ is real parameter.
If $z = x + iy$ is the complex number then $x$ is the real part of $z$ and $y$ is the imaginary part of $z$. Here, $i$ is the imaginary number.
Determine the real part and imaginary part of $z = 1 - t + i\sqrt {{t^2} + t + 2} $ by comparing with $z = x + iy$.
$x = 1 - t$ and $y = \sqrt {{t^2} + t + 2} $
Square both sides of the equation$y = \sqrt {{t^2} + t + 2} $.
${y^2} = {t^2} + t + 2$
${y^2} = {t^2} + t + 1 + 1$
Use the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ to write the term${t^2} + t + 1$.
${y^2} = {\left( {t + 1} \right)^2} + 1 \ldots (1)$
Since,$x = 1 - t$, we write $t = 1 - x$.
Substitute $t = 1 - x$ into the equation $(1)$.
${y^2} = {\left( {1 - x + 1} \right)^2} + 1$
$ \Rightarrow {y^2} = {\left( {1 - x + 1} \right)^2} + 1$
$ \Rightarrow {y^2} = {\left( {2 - x} \right)^2} + 1$
$ \Rightarrow {y^2} - {\left( {2 - x} \right)^2} = 1$
The standard form of hyperbola with center at $(0,0)$ and transverse axis on $y$- axis is given as;
$\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$
Comparing ${y^2} - {\left( {2 - x} \right)^2} = 1$ with the standard form $\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$;
We find the equation ${y^2} - {\left( {2 - x} \right)^2} = 1$ is the equation of hyperbola with center at $(2,0)$ where $a = 1$ and $b = 1$

Option A hyperbola is the correct answer.

Note: The X-axis and Y-axis in two-dimensional geometry, there are two axes in the Argand plane.
The axis which is horizontal is called the real axis
The axis which is vertical is called the imaginary axis
The real part doesn’t contain $i$ term and the imaginary part is the real number attached with the $i$ imaginary number.
Imaginary number is ,$i = \sqrt 1 $ .
The standard form of hyperbola with center at $(0,0)$ and transverse axis on $y$- axis is given as;
$\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$
And
The standard form of hyperbola with center at $(0,0)$ and transverse axis on $x$- axis is given as;
$\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$