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Question: Let \(z = 1 - t + i\sqrt {{t^2} + t + 2} \), where \(t\) is the real parameter. The locus of \[z\] i...

Let z=1−t+it2+t+2z = 1 - t + i\sqrt {{t^2} + t + 2} , where tt is the real parameter. The locus of zz in the argand plane is:
A. A hyperbola
B. An ellipse
C. A straight line
D. None of these

Explanation

Solution

This is a question of complex analysis. If z=x+iyz = x + iy is the complex number then xx is called as the real part of zz and yy is called as the imaginary part of zz. Here, ii 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, xx-axis is the real axis and yy is the imaginary axis.
Compare z=1−t+it2+t+2z = 1 - t + i\sqrt {{t^2} + t + 2} with the complex numberz=x+iyz = x + iy.
Where x=1−tx = 1 - t and y=t2+t+2y = \sqrt {{t^2} + t + 2} .
Remove the real parameter tt with the help of x=1−tx = 1 - t and y=t2+t+2y = \sqrt {{t^2} + t + 2} . The equation in terms of xxand yy 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+it2+t+2z = 1 - t + i\sqrt {{t^2} + t + 2} , where tt is real parameter.
If z=x+iyz = x + iy is the complex number then xx is the real part of zz and yy is the imaginary part of zz. Here, ii is the imaginary number.
Determine the real part and imaginary part of z=1−t+it2+t+2z = 1 - t + i\sqrt {{t^2} + t + 2} by comparing with z=x+iyz = x + iy.
x=1−tx = 1 - t and y=t2+t+2y = \sqrt {{t^2} + t + 2}
Square both sides of the equationy=t2+t+2y = \sqrt {{t^2} + t + 2} .
y2=t2+t+2{y^2} = {t^2} + t + 2
y2=t2+t+1+1{y^2} = {t^2} + t + 1 + 1
Use the algebraic identity (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} to write the termt2+t+1{t^2} + t + 1.
y2=(t+1)2+1…(1){y^2} = {\left( {t + 1} \right)^2} + 1 \ldots (1)
Since,x=1−tx = 1 - t, we write t=1−xt = 1 - x.
Substitute t=1−xt = 1 - x into the equation (1)(1).
y2=(1−x+1)2+1{y^2} = {\left( {1 - x + 1} \right)^2} + 1
⇒y2=(1−x+1)2+1\Rightarrow {y^2} = {\left( {1 - x + 1} \right)^2} + 1
⇒y2=(2−x)2+1\Rightarrow {y^2} = {\left( {2 - x} \right)^2} + 1
⇒y2−(2−x)2=1\Rightarrow {y^2} - {\left( {2 - x} \right)^2} = 1
The standard form of hyperbola with center at (0,0)(0,0) and transverse axis on yy- axis is given as;
y2a2−x2b2=1\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1
Comparing y2−(2−x)2=1{y^2} - {\left( {2 - x} \right)^2} = 1 with the standard form y2a2−x2b2=1\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1;
We find the equation y2−(2−x)2=1{y^2} - {\left( {2 - x} \right)^2} = 1 is the equation of hyperbola with center at (2,0)(2,0) where a=1a = 1 and b=1b = 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 ii term and the imaginary part is the real number attached with the ii imaginary number.
Imaginary number is ,i=1i = \sqrt 1 .
The standard form of hyperbola with center at (0,0)(0,0) and transverse axis on yy- axis is given as;
y2a2−x2b2=1\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1
And
The standard form of hyperbola with center at (0,0)(0,0) and transverse axis on xx- axis is given as;
x2a2−y2b2=1\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1