Question:
If $\left|\frac{z-25}{z-1}\right|=5$ , the value of |z|
If $\left|\frac{z-25}{z-1}\right|=5$ , the value of |z|
Updated On: Jun 18, 2022
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The Correct Option is C
Solution and Explanation
Given that
$\left|\frac{z-25}{z-1}\right|=5 \Rightarrow\left|z-25\right|=5\left|z-1\right|$
Let Z = x + iy, then
$\left|x+iy-25\right|=5\left|x+iy-1\right|$
$\Rightarrow \left|\left(x-25\right)+iy\right|=5\left|x-1+iy\right|$
Squaring both sides, we get
$\left(x-25\right)^{2}+y^{2}=25 \left\{\left(x-1\right)^{2}+y^{2}\right\}$
$\Rightarrow x^{2}-50x+625+y^{2} $
$\quad\quad\quad\quad\quad\quad=25x^{2} -50x+25+25y^{2}$
$\Rightarrow24x^{2}+24 y^{2}-600=0$
$\Rightarrow x^{2}+y^{2}-25=0$
$\Rightarrow\left|x+iy\right|^{2}=25 \Rightarrow \left|Z\right|^{2}=5^{2}$
$\Rightarrow\left|Z\right|=5$
$\left|\frac{z-25}{z-1}\right|=5 \Rightarrow\left|z-25\right|=5\left|z-1\right|$
Let Z = x + iy, then
$\left|x+iy-25\right|=5\left|x+iy-1\right|$
$\Rightarrow \left|\left(x-25\right)+iy\right|=5\left|x-1+iy\right|$
Squaring both sides, we get
$\left(x-25\right)^{2}+y^{2}=25 \left\{\left(x-1\right)^{2}+y^{2}\right\}$
$\Rightarrow x^{2}-50x+625+y^{2} $
$\quad\quad\quad\quad\quad\quad=25x^{2} -50x+25+25y^{2}$
$\Rightarrow24x^{2}+24 y^{2}-600=0$
$\Rightarrow x^{2}+y^{2}-25=0$
$\Rightarrow\left|x+iy\right|^{2}=25 \Rightarrow \left|Z\right|^{2}=5^{2}$
$\Rightarrow\left|Z\right|=5$
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Concepts Used:
Complex Number
A Complex Number is written in the form
a + ib
where,
- “a” is a real number
- “b” is an imaginary number
The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.