If t and c are two complex numbers such that -t- -c- -t-1 and z-at-b-t-c- z-x-iy- Locus of z is where a- b are complex numbers

The correct option is C Circle z=at+bt c zt zc=at+b t(z a)=b+zc t=b+zcz a |t|=|b+zcz a|=1 |z+bcz a|=1|c| Now |c|≠1 HenceLocus ...

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Byju's Answer

Standard XII

Mathematics

Geometrical Representation of Argument and Modulus

If t and c ar...

Question

If t and c are two complex numbers such that $| t | \neq | c |, | t | = 1$ and $z = (a t + b) / (t - c), z = x + i y$ . Locus of z is (where a, b are complex numbers)

Line segment

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Straight line

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Circle

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None of these

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Solution

The correct option is C Circle
$z = \frac{a t + b}{t - c}$
$z t - z c = a t + b$
$t (z - a) = b + z c$
$t = \frac{b + z c}{z - a}$
$| t | = | \frac{b + z c}{z - a} | = 1$
$| \frac{z + \frac{b}{c}}{z - a} | = \frac{1}{| c |}$
Now $| c | \neq 1$
Hence
Locus of z is a circle.

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Which of the following is only CORRECT combination?

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If t and c are two complex numbers such that |t|≠|c|,|t|=1 and z=(at+b)/(t−c),z=x+iy. Locus of z is (where a, b are complex numbers)

The correct option is C Circlez=at+bt−czt−zc=at+bt(z−a)=b+zct=b+zcz−a|t|=|b+zcz−a|=1|z+bcz−a|=1|c|Now |c|≠1HenceLocus of z is a circle.

If t and c are two complex numbers such that $| t | \neq | c |, | t | = 1$ and $z = (a t + b) / (t - c), z = x + i y$ . Locus of z is (where a, b are complex numbers)

The correct option is C Circle
$z = \frac{a t + b}{t - c}$
$z t - z c = a t + b$
$t (z - a) = b + z c$
$t = \frac{b + z c}{z - a}$
$| t | = | \frac{b + z c}{z - a} | = 1$
$| \frac{z + \frac{b}{c}}{z - a} | = \frac{1}{| c |}$
Now $| c | \neq 1$
Hence
Locus of z is a circle.