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If Z − 1 Z + 1 is Purely Imaginary Number ( Z ≠ − 1 ), Find the Value of | Z | . - Mathematics

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प्रश्न

If \[\frac{z - 1}{z + 1}\] is purely imaginary number (\[z \neq - 1\]), find the value of \[\left| z \right|\].

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उत्तर

Let \[z = x + iy\]

Then,  

\[\frac{z - 1}{z + 1} = \frac{x + iy - 1}{x + iy + 1}\]

\[ = \frac{\left( x - 1 \right) + iy}{\left( x + 1 \right) + iy} \times \frac{\left( x + 1 \right) - iy}{\left( x + 1 \right) - iy}\]

\[ = \frac{x^2 + x - ixy - x - 1 + iy + ixy + iy - i^2 y^2}{\left( x + 1 \right)^2 - i^2 y^2}\]

\[ = \frac{x^2 + y^2 - 1 + 2iy}{x^2 + 1 + 2x + y^2} [ \because i^2 = - 1]\]

If \[\frac{z - 1}{z + 1}\] is purely imaginary number, then

\[\text { Re }\left( \frac{z - 1}{z + 1} \right) = 0\]

\[ \Rightarrow x^2 + y^2 - 1 = 0\]

\[ \Rightarrow x^2 + y^2 = 1\]

\[ \Rightarrow \left| z \right|^2 = 1\]

\[ \Rightarrow \left| z \right| = 1\]

Thus, the value of \[\left| z \right|\] is 1.

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पाठ 13: Complex Numbers - Exercise 13.2 [पृष्ठ ३३]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 13 Complex Numbers
Exercise 13.2 | Q 20 | पृष्ठ ३३

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