Find the locus of a complex number, \[z = x + iy\], satisfying the relation \[\left| {\dfrac{{z - 3i}}{{z + 3i}}} \right| \leqslant \sqrt 2 \]. Illustrate the locus of \[z\] in the Argand plane.
Answer
612.9k+ views
Hint: A complex number is a number that can be expressed in the form \[p + iq\]
where,
p and q are real numbers and ‘i’ is a solution of the equation
\[{x^2} = - 1\]
\[\sqrt { - 1} = i\] or \[{i^2} = - 1\]
General form of complex number
\[z = p + iq\]
where,
p is known as the real part denote by \[\operatorname{Re} z\]
q is known as the imaginary part denote by \[im.z.\]
The modulus and conjugate of complex numbers-
Let \[z = m + in\] be a complex number.
Then, the modulus of z, denotes by \[\left| z \right| = \sqrt {{m^2} - {n^2}} \] and
The conjugate of z, denoted by \[\widetilde z\] is the complex number \[m - ni\]
In the Argand plane the modulus of the complex number \[m + ni = \sqrt {{m^2} - {n^2}} \] is the distance between the point \[(m,n)\] and the origin \[(0,0)\]
The x-axis termed as real axis and the y-axis termed as imaginary axis.
Locus complex number is obtained by letting
\[z = x + yi\] and simplifying the expression.
Operation of modulus, conjugate pairs and arguments are to be used for determining the locus of complex numbers.
Therefore,
Complete step by step answer:
Let \[z = x + iy\]
Then given \[\left| {\dfrac{{z - 3i}}{{z + 3i}}} \right| \leqslant \sqrt 2 - - - - - 1.\]
Putting the value of z in equation 1.
\[\left| {\dfrac{{x + iy - 3i}}{{x + iy + 3i}}} \right| \leqslant \sqrt 2 \]
\[\left| {\dfrac{{x + i(y - 3)}}{{x + i(y + 3)}}} \right| \leqslant \sqrt 2 \]
\[\sqrt {{x^2} + {{(y - 3)}^2}} \leqslant \sqrt 2 \sqrt {{x^2} + {{(y + 3)}^2}} \]
On squaring on both sides we get
\[{x^2} + {(y - 3)^2} \leqslant ({x^2} + {(y + 3)^2})\]
\[{x^2} + {y^2} + 9 - 6y \leqslant 2{x^2} + 2{y^2} + 18 + 12y\]
Subtracting like terms
\[{x^2} + {y^2} + 18y + 9 \geqslant 0\]
A circle with centre \[(0, - 9)\]and radius \[6\sqrt 2 \] unit
Note: Equation of complex locus of circle
The locus of z that satisfies the equation \[\left| {z - zo} \right| = r\]
Where, \[zo\] is a fixed complex number and r is a fixed positive real number consists of all points \[z\] whose distance from \[zo\] is \[r\].
Therefore, \[\left| {z - zo} \right| = r\] is the complex form of the equation of a circle.
\[\left| {z - zo} \right| < r\]represents the point interior of the circle.
\[\left| {z - zo} \right| > r\]represents the points exterior of the circle.
where,
p and q are real numbers and ‘i’ is a solution of the equation
\[{x^2} = - 1\]
\[\sqrt { - 1} = i\] or \[{i^2} = - 1\]
General form of complex number
\[z = p + iq\]
where,
p is known as the real part denote by \[\operatorname{Re} z\]
q is known as the imaginary part denote by \[im.z.\]
The modulus and conjugate of complex numbers-
Let \[z = m + in\] be a complex number.
Then, the modulus of z, denotes by \[\left| z \right| = \sqrt {{m^2} - {n^2}} \] and
The conjugate of z, denoted by \[\widetilde z\] is the complex number \[m - ni\]
In the Argand plane the modulus of the complex number \[m + ni = \sqrt {{m^2} - {n^2}} \] is the distance between the point \[(m,n)\] and the origin \[(0,0)\]
The x-axis termed as real axis and the y-axis termed as imaginary axis.
Locus complex number is obtained by letting
\[z = x + yi\] and simplifying the expression.
Operation of modulus, conjugate pairs and arguments are to be used for determining the locus of complex numbers.
Therefore,
Complete step by step answer:
Let \[z = x + iy\]
Then given \[\left| {\dfrac{{z - 3i}}{{z + 3i}}} \right| \leqslant \sqrt 2 - - - - - 1.\]
Putting the value of z in equation 1.
\[\left| {\dfrac{{x + iy - 3i}}{{x + iy + 3i}}} \right| \leqslant \sqrt 2 \]
\[\left| {\dfrac{{x + i(y - 3)}}{{x + i(y + 3)}}} \right| \leqslant \sqrt 2 \]
\[\sqrt {{x^2} + {{(y - 3)}^2}} \leqslant \sqrt 2 \sqrt {{x^2} + {{(y + 3)}^2}} \]
On squaring on both sides we get
\[{x^2} + {(y - 3)^2} \leqslant ({x^2} + {(y + 3)^2})\]
\[{x^2} + {y^2} + 9 - 6y \leqslant 2{x^2} + 2{y^2} + 18 + 12y\]
Subtracting like terms
\[{x^2} + {y^2} + 18y + 9 \geqslant 0\]
A circle with centre \[(0, - 9)\]and radius \[6\sqrt 2 \] unit
Note: Equation of complex locus of circle
The locus of z that satisfies the equation \[\left| {z - zo} \right| = r\]
Where, \[zo\] is a fixed complex number and r is a fixed positive real number consists of all points \[z\] whose distance from \[zo\] is \[r\].
Therefore, \[\left| {z - zo} \right| = r\] is the complex form of the equation of a circle.
\[\left| {z - zo} \right| < r\]represents the point interior of the circle.
\[\left| {z - zo} \right| > r\]represents the points exterior of the circle.
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