Question

Find real and imaginary parts of the complex number \[z = \dfrac{{3{i^{20}} - {i^{19}}}}{{2i - 1}}\].

Answer 1

Hint: We convert the power of iota (i) in the lowest form. Thereafter, we rationalise the denominator of iota(i).

Formula used: \[{(i)^4} = 1,\,\,{i^2} = - 1,\,\,{i^3} = - i\]
\[z = a + ib\] (standard form of complex Number), with the real part and the imaginary part.

Complete step by step answer:
(1) We have, \[z = \dfrac{{3{{(i)}^{20}} - {i^{19}}}}{{2i - 1}}\]
\[ \Rightarrow z = \dfrac{{3{{({i^4})}^5} - {{({i^4})}^4}\left( i \right){{(i)}^2}}}{{2i - 1}}\]
\[ \Rightarrow z = \dfrac{{3{{(1)}^5} - {{(1)}^4} \times (i){{( - 1)}^2}}}{{2i - 1}}\]
\[ \Rightarrow z = \dfrac{{3 + i}}{{2i - 1}}\,\,\,\,\,\,\,(\because {i^4} = 1)\]
(2) In step 1 simplification, we see that iota(i) comes in denominator.
Therefore on rationalize,
\[z = \dfrac{{3 + i}}{{2i - 1}} \times \dfrac{{2i + 1}}{{2i + 1}}\] ,\[z = \dfrac{{(3 + i)(2i + 1)}}{{(2i - 1)(2i + 1)}}\]
Using algebraic identity \[(a + b)(a - b) = {a^2} - {b^2}\] in denominator
\[z = \dfrac{{3(2i + 1) + i(2i + 1)}}{{{{(2i)}^2} - {{(1)}^2}}}\]
\[ = \dfrac{{6i + 3 + 2{{(i)}^2} + i}}{{4{{(i)}^2} - 1}}\]
$\dfrac{{6i + 3 + 2( - 1) + i}}{{4( - 1) - 1}}$
\[ = \dfrac{{6i + 3 - 2 + 1i}}{{4( - 1) - 1}}\]
\[ = \dfrac{{7i + 1}}{{ - 4 - 1}}\]
\[ = \dfrac{{7i + 1}}{{ - 5}}\]
\[z = \dfrac{{ - 1}}{5} - i\dfrac{7}{5}\]
(4) \[z = \dfrac{{ - 1}}{5} - i\dfrac{7}{5}\] is the same form as \[z = a + ib\].
Hence, \[z = \dfrac{{ - 1}}{5} - i\dfrac{7}{5}\] is the required standard form.
(5) On comparing \[z = \dfrac{{ - 1}}{5} - i\dfrac{7}{5}\] with \[z = a + ib\] we have
\[a = \dfrac{{ - 1}}{5}\,\,and\,\,b = - \dfrac{7}{5}\].
Where is called real part of the complex number and is called imaginary part of complex number

Hence, real part \[ = \dfrac{1}{5}\], Imaginary part $ = - \dfrac{7}{5}$

Additional Information: A complex number is a number that can be expressed in the form \[a{\text{ }} + {\text{ }}bi,\,\]where \[a{\text{ }}and{\text{ }}b\]are real numbers, and \[i\]represents the imaginary unit, satisfying the equation \[{i^2} = - 1\]because no real number satisfies this equation, $i$ is called an imaginary number.

Note: When we are comparing real and imaginary parts then in standard complex numbers, imaginary part is the coefficient of and the part without iota(i) is called real.