Question

Let $z \ne - i$ be any complex number such that $\dfrac{{z - i}}{{z + i}}$ is a purely imaginary number. Then $z + \dfrac{1}{z}$ is
A. 0
B. Any non-zero real number other than 1
C. Any non-zero real number
D. Any purely imaginary number

Answer 1

We assume $z = x + iy$ and substitute it in fraction which is given as purely imaginary. Solve this fraction by rationalizing it and then in the end take the real part as zero. This will give us the relation between x and y and then we can write the value of z and find the value of $z + \dfrac{1}{z}$

Complete step-by-step answer:
Let us assume $z = x + iy$
Substitute the value of z in fraction $\dfrac{{z - i}}{{z + i}}$we get
$\Rightarrow \dfrac{{x + iy - i}}{{x + iy + i}}$
Take common I from the imaginary terms and write the fraction
$\Rightarrow \dfrac{{x + i(y - 1)}}{{x + i(y + 1)}}$
Now we rationalize by multiplying both numerator and denominator by $x - i(y + 1)$

$$\Rightarrow \dfrac{{x + i(y - 1)}}{{x + i(y + 1)}} \times \dfrac{{x - i(y + 1)}}{{x - i(y + 1)}} \\\ \Rightarrow \dfrac{{x[x - i(y + 1)] + i(y - 1)[x - i(y + 1)]}}{{[x + i(y + 1)][x - i(y + 1)]}} \\\$$

Using the formula $(a + b)(a - b) = {a^2} - {b^2}$in the denominator
$\Rightarrow \dfrac{{{x^2} - ix(y + 1) + ix(y - 1) - {i^2}(y - 1)(y + 1)}}{{{x^2} - {{[i(y + 1)]}^2}}}$
Taking common ix in the numerator
$\Rightarrow \dfrac{{{x^2} + ix[(y - 1) - (y + 1)] - {i^2}[(y - 1)(y + 1)]}}{{{x^2} - {i^2}{{(y + 1)}^2}}}$
Using the formula $(a + b)(a - b) = {a^2} - {b^2}$ in the numerator and substituting the value of ${i^2} = - 1$ in both numerator and denominator.

$$\Rightarrow \dfrac{{{x^2} + ix[y - 1 - y - 1] - ( - 1)[{y^2} - {1^2}]}}{{{x^2} - ( - 1){{(y + 1)}^2}}} \\\ \Rightarrow \dfrac{{{x^2} + ix( - 2) + ({y^2} - 1)}}{{{x^2} + {{(y + 1)}^2}}} \\\$$

Writing the real and imaginary terms separately
$\Rightarrow \dfrac{{({x^2} + {y^2} - 1) - 2ix}}{{{x^2} + {{(y + 1)}^2}}}$
Now we can write the real and imaginary part separately
$\Rightarrow \dfrac{{{x^2} + {y^2} - 1}}{{{x^2} + {{(y + 1)}^2}}} - i\dfrac{{2x}}{{{x^2} + {{(y + 1)}^2}}}$
Since the number is pure imaginary, so the real term is zero
$\Rightarrow \dfrac{{{x^2} + {y^2} - 1}}{{{x^2} + {{(y + 1)}^2}}} = 0$
Cross multiplying the values
$\Rightarrow {x^2} + {y^2} - 1 = 0$
Shifting the constant values to right side of the equation
$\Rightarrow {x^2} + {y^2} = 1$ … (1)
This is an equation of circle with center $(0,0)$ and radius 1
Solving $z + \dfrac{1}{z}$by substituting $z = x + iy$
$\Rightarrow x + iy + \dfrac{1}{{x + iy}}$
Rationalizing the fraction part by multiplying both numerator and denominator

$$\Rightarrow x + iy + \dfrac{1}{{x + iy}} \times \dfrac{{x - iy}}{{x - iy}} \\\ \Rightarrow x + iy + \dfrac{{x - iy}}{{(x - iy)(x + iy)}} \\\$$

Using the formula $(a + b)(a - b) = {a^2} - {b^2}$ in the denominator

$$\Rightarrow x + iy + \dfrac{{x - iy}}{{{{(x)}^2} - {{(iy)}^2}}} \\\ \Rightarrow x + iy + \dfrac{{x - iy}}{{{x^2} - {i^2}{y^2}}} \\\$$

Substitute the value of ${i^2} = - 1$

$$\Rightarrow x + iy + \dfrac{{x - iy}}{{{x^2} - ( - 1){y^2}}} \\\ \Rightarrow x + iy + \dfrac{{x - iy}}{{{x^2} + {y^2}}} \\\$$

Substitute the value of ${x^2} + {y^2} = 1$ from equation (i) in the denominator

$$\Rightarrow x + iy + \dfrac{{x - iy}}{1} \\\ \Rightarrow x + iy + x - iy \\\ \Rightarrow 2x \\\$$

We write the value as $2x + 0i$. So the imaginary part is zero.
Thus the value of $z + \dfrac{1}{z}$ is purely real and can take any value.

So, the correct answer is “Option C”.

Note:
Alternative method:
We have $z = x + iy$
and we know $\overline z = \dfrac{1}{z}$
So we can write $\dfrac{1}{z} = \overline z = \overline {x + iy} = x - iy$

$$\Rightarrow z + \dfrac{1}{z} = x + iy + x - iy \\\ \Rightarrow z + \dfrac{1}{z} = 2x \\\$$

So, the value of $z + \dfrac{1}{z}$ has no imaginary part. So, the value can be any real number.
So, option C is correct.