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If \[z = (\lambda + 3) + i\sqrt {5 - {\lambda ^2}} \] then the locus of z is a
A) A circle
B) A straight line
C) A parabola
D) None of these


Answer
VerifiedVerified
164.1k+ views
Hint: in this question we have to find what shape locus of the point in the complex plane represent. First, write the given complex number as a combination of real and imaginary numbers. Put z in form of real and imaginary number into the equation

Formula Used:\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number
i is iota
Square of iota is equal to the negative of one



Complete step by step solution:Given: A complex number \[z = (\lambda + 3) + i\sqrt {5 - {\lambda ^2}} \]
Now we have complex number \[z = (\lambda + 3) + i\sqrt {5 - {\lambda ^2}} \]
We know that complex number is written as a combination of real and imaginary number.
\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number

iy is a imaginary part of complex number

Put this value in\[z = (\lambda + 3) + i\sqrt {5 - {\lambda ^2}} \]
\[x + iy = (\lambda + 3) + i\sqrt {5 - {\lambda ^2}} \]
Compare both sides
\[x = (\lambda + 3)\]
\[y = \sqrt {5 - {\lambda ^2}} \]
\[{y^2} = 5 - {\lambda ^2}\]
\[5 - {y^2} = {\lambda ^2}\]……………………………………….. (i)
\[x - 3 = \lambda \]
\[{(x - 3)^2} = {\lambda ^2}\]……………………………………. (ii)
From equation (i) and (ii)
\[{(x - 3)^2} = 5 - {y^2}\]
\[{(x - 3)^2} + {y^2} = 5\]
It is in the form of \[{{\bf{x}}^{\bf{2}}}\; + {\rm{ }}{{\bf{y}}^{\bf{2}}}\; + {\rm{ }}{\bf{2gx}}{\rm{ }} + {\rm{ }}{\bf{2fy}}{\rm{ }} + {\rm{ }}{\bf{c}}{\rm{ }} = {\rm{ }}{\bf{0}}\]
This equation represents the circle
Here \[{(x - 3)^2} + {y^2} = 5\]represent the equation of circle therefore locus of point represent circle.



Option ‘A’ is correct



Note: Complex number is a number which is a combination of real and imaginary number. So in complex number questions, we have to represent number as a combination of real and its imaginary part. Imaginary part is known as iota. Square of iota is equal to negative one.