Question
Question: If \( \left| z \right| = 1, \) prove that \( \dfrac{{z - 1}}{{z + 1}}(z \ne - 1), \) is a pure imagi...
If ∣z∣=1, prove that z+1z−1(z=−1), is a pure imaginary number. What will you conclude if z=1 ?
Solution
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 l2=−1 the square of an imaginary number b1 is −b2 , 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 ∣z∣2=x2+y2 therefore, the condition ∣z∣=1 is equivalent to,
We know that ∣z∣=1
⇒x2+y2=1
Now, according to the equation
z+1z−1=x+iy+1x+iy−1
Separating both side
=(x+1+iy)(x+1−iy)(x−1+iy)(x+1−iy) =(x+1)2+y2(x2+y2−1)+2iy
=(x+1)2+y22iy
Hence, z+1z−1 is purely imaginary
When ∣z∣=1
Provided z=−1
when z=1 ,
we have z+1z−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 z+1z−1 is a pure imaginary number.
So, the correct answer is “z+1z−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 i2=−1 . The square of the imaginary number of bi is −b2 . Imaginary numbers are also called complex numbers, and are used in real life applications, such as electricity, as well as quadratic equations.