Question

Find the modulus of the complex number$\dfrac{i}{{1 - i}}$.

Answer 1

Hint: Express $\dfrac{i}{{1 - i}}$ in the form of z = x + iy where x and y are real numbers, by multiplying the numerator and denominator by the complex conjugate of the denominator.
Then use the formula \[|z| = |x + iy| = \sqrt {{x^2} + {y^2}} \] to compute the answer.

Complete step by step answer:
We have a complex fraction with us: $\dfrac{i}{{1 - i}}$
Let’s begin by recalling the definition of the modulus of a complex number.
Consider a complex number z = x + iy where x and y are real numbers and $i = \sqrt { - 1} $.
Here x and y are called the real and imaginary parts of the complex number z.
Then the modulus of the complex number z is given by \[|z| = |x + iy| = \sqrt {{x^2} + {y^2}} \]
Now, let us go back to the given complex number $\dfrac{i}{{1 - i}}$.
We could have applied the definition to the given complex number if it was also expressed in the form of z = x + iy where x and y are real numbers.
Because we need the real and imaginary part of a complex number to compute the modulus.
So, we will do some operations on the given number without affecting the original value and express it in the required format.
Let $z = \dfrac{i}{{1 - i}}$.
We know that the complex conjugate of z = x + iy is $\overline z = x - iy$ and the complex conjugate of z = x - iy
Therefore, the complex conjugate of $1 - i$ is $1 + i$.
Now, multiply $1 + i$ with the numerator and the denominator of $z = \dfrac{i}{{1 - i}}$.
This is done as follows:
$
  z = \dfrac{i}{{1 - i}} \\
   \Rightarrow z = \dfrac{i}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}} \\
 $
We will use the identity $z\overline z = (a + bi)(a - bi) = {a^2} + {b^2}$ for z = a + bi in the denominator.
Also, in the numerator we will use the fact that \[{i^2} = - 1\]
$
  z = \dfrac{i}{{1 - i}} \\
   \Rightarrow z = \dfrac{i}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}} = \dfrac{{i + {i^2}}}{{{1^2} + {1^2}}} = \dfrac{{i - 1}}{2} \\
 $
Now, the denominator of $z = \dfrac{i}{{1 - i}}$ is a real number which is 2.
So, we can now express z in the form of x + iy as follows:

We can see that \[\dfrac{{( - 1)}}{2}\] and \[\dfrac{1}{2}\] are the real and imaginary parts respectively of $z = \dfrac{i}{{1 - i}}$.
Now that we have expressed the given complex number in the form of x + iy, we can compute its modulus.
Therefore, modulus of $z = \dfrac{i}{{1 - i}}$ is
\[|z| = |\dfrac{{( - 1)}}{2} + i\dfrac{1}{2}| = \sqrt {{{(\dfrac{{ - 1}}{2})}^2} + {{(\dfrac{1}{2})}^2}} = \sqrt {\dfrac{1}{4} + \dfrac{1}{4}} = \sqrt {\dfrac{2}{4}} = \dfrac{{\sqrt 2 }}{2}\]or$\dfrac{1}{{\sqrt 2 }}$

Hence the modulus of $\dfrac{i}{{1 - i}}$ is $\dfrac{1}{{\sqrt 2 }}$.

Note: Please note that in the conjugate of a complex number, the sign of the imaginary part of the complex number should be changed.
Therefore, one should always check if the complex number is given in the form of \[iy{\text{ }} + {\text{ }}x\] or \[x{\text{ }} + {\text{ }}iy\]. Then the complex conjugate would be \[ - iy{\text{ }} + {\text{ }}x\] or \[x{\text{ }} - {\text{ }}iy\] accordingly.