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# 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.

Last updated date: 13th Jun 2024
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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.