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प्रश्न
If 2 = x + iy is a complex number such that `|(z - 4"i")/(z + 4"i")|` = 1 show that the locus of z is real axis
उत्तर
`|(z - 4"i")/(z + 4"i")|` = 1
⇒ |z – 4i| = |z + 4i|
Let z = x + iy
⇒ |x + iy – 4i| = |x + iy + 4i|
⇒ |x + i(y – 4)| = |x +(y + 4)|
⇒ `sqrt(x^2 + (y - 4)^2`
= `sqrt(x^2 + (y + 4)^2`
Squaring on both sides, we get
x2 + y2 – 8y + 16 = x2 + y2 + 16 + 8y
⇒ – 16y = 0
⇒ y = 0 in two equation of real axis.
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