Question

Real and imaginary parts of complex number $\dfrac{{1 + i}}{{1 - i}}$ respectively are:
$A)1,1$
$B)0,0$
$C)0,1$
$D)1,0$

Answer 1

Hint: Since the given question is in the form of complex numbers because it contains both the real number system and imaginary terms also. Complex numbers can be expressed as in the form of $z = x + iy$ where $x,y$ are the real numbers and $i$ is the imaginary term.

Complete step-by-step solution:
Since the complex numbers are in the form of division of the two complex terms, like $\dfrac{{1 + i}}{{1 - i}}$ and we need to find its real and imaginary parts separately.
To solve the given division format problem, we will use the concept of the conjugation method. Which is the process of the complex numbers representing the reflection of that complex number about the imaginary axis of the argument plane. Like $i$ of the complex number is replaced with $ - i$ and thus which is the complex conjugate process. And this process can be applied for the denominator terms if the given question is in division
Hence using this process, we will apply the conjugate of the given numbers by just multiplying and dividing the terms by the value $1 + i$ (because which is the conjugate complex of the denominator term $1 - i$)
Therefore, we get $\dfrac{{1 + i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}$ since $\dfrac{{1 + i}}{{1 + i}} = 1$ so the values would not affect after applying the addition terms on the given question.
Further solving we get $\dfrac{{1 + i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}} = \dfrac{{{{(1 + i)}^2}}}{{{1^2} - {i^2}}}$
Thus, we have $\dfrac{{{{(1 + i)}^2}}}{{{1^2} - {i^2}}} = \dfrac{{1 + {i^2} + 2i}}{{1 - {i^2}}}$ since we know that ${i^2} = - 1$
Hence, we have $\dfrac{{1 + {i^2} + 2i}}{{1 - {i^2}}} = \dfrac{{1 - 1 + 2i}}{{1 + 1}} = \dfrac{{2i}}{2}$
Therefore, we get $i$ as the simplified answer and which can be rewritten in the complex form as $0 + 1i$
Hence there is no real part and thus we get real part as $0$ and the imaginary part as $1$
Hence the option $C)0,1$ is correct.

Note: Note that complex imaginary values $i$ can be reframed in the real form of ${i^2} = - 1$ or in the inverse form as $\sqrt { - 1} = i$
Also note that the imaginary value contained number is called as the imaginary part like $ - i$ is the form $ - 1(i)$ of imaginary part