Questions & Answers

Question

Answers

(A). Line segment

(B). Straight line

(C). Circle

(D). None of these

Answer
Verified

Hint: First write the condition given out the complex number t. As you know the value of $\left| t \right|$, you have to bring an equation for t. First do cross multiplication for a given relation. Now you have variable t on both sides. Try to manipulate and bring all terms with t on to the left-hand side. Now get the value of t. Substitute known values and then try to get remaining locus.

__Complete step-by-step solution -__

Given condition on the complex number t is written as:

$\left| t \right| = 1$ ………………(1)

Given condition on the complex number c is written as:

$\left| c \right| \ne \left| t \right|$

By using equation (1) we can say that $\left| c \right|$ can be written as:

$\left| c \right| \ne 1$ ………………(2)

Given relation between z,c,t,a,b can be written in form of:

$z = \dfrac{{at + b}}{{t - c}}$

By cross multiplying the terms in above equation, we can get it as:

$z\left( {t - c} \right) = at + b$

Multiply term inside bracket with z and remove bracket, we get:

$zt - zc = at + b$

By subtracting with at on both sides, we get it as:

$zt - zc - at = at + b - at$

By simplifying the above equation, we can write it as:

$zt - at - zc = 0 + b$

By adding zc on both sides, we get it as:

$zt - at - zc + zc = zc + b$

By cancelling the common term on both sides, we get it as:

$zt - at = zc + b$

By taking t as common on left, c as common on right, we get:

$\left( {z - a} \right)t = \left( {z + \dfrac{b}{c}} \right)c$

By dividing with $\left( {z - a} \right).c$ on both sides, we get it as:

$\dfrac{{\left( {z - a} \right)t}}{{\left( {z - a} \right)c}} = \dfrac{{\left( {z + \dfrac{b}{c}} \right)c}}{{\left( {z - a} \right)c}}$

By cancelling common terms and applying magnitude, we get:

$\left| {\dfrac{t}{c}} \right| = \left| {\dfrac{{z + \dfrac{b}{c}}}{{z - a}}} \right|$

By substituting equation (1), we get the equation in form:

$\dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + \dfrac{b}{c}}}{{z - a}}} \right|$

By evaluating (2) we know $\left| c \right| \ne 1$. So when we cross multiply as $z = x + iy$, we get coefficients of ${x^2},{y^2}$ as 1 on the left hand side but on the right hand side $ \ne 1$. So, the final equation will have terms of${x^2},{y^2}$. But the magnitude of the complex number does not cancel.

So, we know $\left| {\dfrac{{z + \alpha }}{{z - \beta }}} \right| = a$ where $a \ne 1$ the locus of z is a circle.

Therefore, option (c) is the correct answer.

Note: It is important that you write $\left| c \right| \ne 1$ as it is the base reason that ${x^2},{y^2}$ terms do not cancel. If they cancel it becomes a straight line which is wrong. So, $\left| c \right| \ne 1$ must be there. Students confuse and forget to multiply z to term c while removing brackets. Be careful at this step because it is the main step.

Given condition on the complex number t is written as:

$\left| t \right| = 1$ ………………(1)

Given condition on the complex number c is written as:

$\left| c \right| \ne \left| t \right|$

By using equation (1) we can say that $\left| c \right|$ can be written as:

$\left| c \right| \ne 1$ ………………(2)

Given relation between z,c,t,a,b can be written in form of:

$z = \dfrac{{at + b}}{{t - c}}$

By cross multiplying the terms in above equation, we can get it as:

$z\left( {t - c} \right) = at + b$

Multiply term inside bracket with z and remove bracket, we get:

$zt - zc = at + b$

By subtracting with at on both sides, we get it as:

$zt - zc - at = at + b - at$

By simplifying the above equation, we can write it as:

$zt - at - zc = 0 + b$

By adding zc on both sides, we get it as:

$zt - at - zc + zc = zc + b$

By cancelling the common term on both sides, we get it as:

$zt - at = zc + b$

By taking t as common on left, c as common on right, we get:

$\left( {z - a} \right)t = \left( {z + \dfrac{b}{c}} \right)c$

By dividing with $\left( {z - a} \right).c$ on both sides, we get it as:

$\dfrac{{\left( {z - a} \right)t}}{{\left( {z - a} \right)c}} = \dfrac{{\left( {z + \dfrac{b}{c}} \right)c}}{{\left( {z - a} \right)c}}$

By cancelling common terms and applying magnitude, we get:

$\left| {\dfrac{t}{c}} \right| = \left| {\dfrac{{z + \dfrac{b}{c}}}{{z - a}}} \right|$

By substituting equation (1), we get the equation in form:

$\dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + \dfrac{b}{c}}}{{z - a}}} \right|$

By evaluating (2) we know $\left| c \right| \ne 1$. So when we cross multiply as $z = x + iy$, we get coefficients of ${x^2},{y^2}$ as 1 on the left hand side but on the right hand side $ \ne 1$. So, the final equation will have terms of${x^2},{y^2}$. But the magnitude of the complex number does not cancel.

So, we know $\left| {\dfrac{{z + \alpha }}{{z - \beta }}} \right| = a$ where $a \ne 1$ the locus of z is a circle.

Therefore, option (c) is the correct answer.

Note: It is important that you write $\left| c \right| \ne 1$ as it is the base reason that ${x^2},{y^2}$ terms do not cancel. If they cancel it becomes a straight line which is wrong. So, $\left| c \right| \ne 1$ must be there. Students confuse and forget to multiply z to term c while removing brackets. Be careful at this step because it is the main step.

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