Question

If $ \left| z \right| = 1, $ prove that $ \dfrac{{z - 1}}{{z + 1}}(z \ne - 1), $ is a pure imaginary number. What will you conclude if $ z = 1 $ ?

Answer 1

Hint: A number that is expressed in terms of the square root of negative number. An imaginary number is a complex number that can be return as a real number multiplied by the imaginary unit which is defined by its property $ {l^2} = - 1 $ the square of an imaginary number $ {b_1} $ is $ - {b^2} $ , for example, $ 5i $ is an imaginary number and its square is $ - 25 $

Complete step-by-step answer:
A part of a conditional statement after then, for example, the conclusion of “if a line is horizontal then the line has slope $ 0 $ is the line that has slope $ 0 $ ”
Let, $ z = x + iy $ then $ {\left| z \right|^2} = {x^2} + {y^2} $ therefore, the condition $ \left| z \right| = 1 $ is equivalent to,
We know that $ \left| z \right| = {1_{}} $
$\Rightarrow {x^2} + {y^2} = 1 $
Now, according to the equation
 $ \dfrac{{z - 1}}{{z + 1}} = \dfrac{{x + iy - 1}}{{x + iy + 1}} $
Separating both side
 $
   = \dfrac{{\left( {x - 1 + iy} \right)\left( {x + 1 - iy} \right)}}{{\left( {x + 1 + iy} \right)\left( {x + 1 - iy} \right)}} \\
   = \dfrac{{({x^2} + {y^2} - 1) + 2iy}}{{{{(x + 1)}^2} + {y^2}}} \\
  $
 $ = \dfrac{{2iy}}{{{{\left( {x + 1} \right)}^2} + {y^2}}} $
Hence, $ \dfrac{{z - 1}}{{z + 1}} $ is purely imaginary
When $ \left| z \right| = 1 $
Provided $ z \ne - 1 $
when $ z = 1 $ ,
we have $ \dfrac{{z - 1}}{{z + 1}} = 0 $
Now, recall that according to the definition that 0 is a pure imaginary number. Since, the point that is imaginary number 0 which corresponds to $ z = 0 $ lies on both the real and imaginary axis. So in this case also $ \dfrac{{z - 1}}{{z + 1}} $ is a pure imaginary number.
So, the correct answer is “$ \dfrac{{z - 1}}{{z + 1}} $ is a pure imaginary number”.

Note: An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit i, which is defined by its property $ {i^2} = - 1 $ . The square of the imaginary number of $ bi $ is $ - {b^2} $ . Imaginary numbers are also called complex numbers, and are used in real life applications, such as electricity, as well as quadratic equations.