Answer
Verified
420.3k+ views
Hint- Here, we will proceed by representing the given complex number in the form of any general complex number i.e., $z = x + iy$ and from there we will find the real and imaginary parts of the given complex number.
Complete step-by-step answer:
Let the given complex number be $z = \dfrac{{2 + i}}{{4i + {{\left( {1 + i} \right)}^2}}}{\text{ }} \to {\text{(1)}}$
As we know that the general form of any complex number is given by $z = x + iy{\text{ }} \to {\text{(2)}}$ where x is the real part of the complex number z and y is the imaginary part of the complex number z.
Now, we will convert the given complex number given by equation (1) in the same form as the general form of any complex number given by equation (2) as under
\[ \Rightarrow z = \dfrac{{2 + i}}{{4i + {1^2} + {i^2} + 2\left( 1 \right)\left( i \right)}}\]
Using ${i^2} = - 1$ in the above equation, we get
\[
\Rightarrow z = \dfrac{{2 + i}}{{4i + 1 + \left( { - 1} \right) + 2i}} \\
\Rightarrow z = \dfrac{{2 + i}}{{6i}} \\
\]
In order to rationalize the above equation, we will multiply and divide the RHS of the above equation by i, we have
i.e., \[
z = \dfrac{{i\left( {2 + i} \right)}}{{i\left( {6i} \right)}} \\
\Rightarrow z = \dfrac{{2i + {i^2}}}{{6{i^2}}} \\
\]
Using ${i^2} = - 1$ in the above equation, we get
\[
\Rightarrow z = \dfrac{{2i + \left( { - 1} \right)}}{{6\left( { - 1} \right)}} \\
\Rightarrow z = - \left( {\dfrac{{2i - 1}}{6}} \right) \\
\Rightarrow z = \dfrac{{1 - 2i}}{6} \\
\Rightarrow z = \dfrac{1}{6} - \dfrac{{2i}}{6} \\
\Rightarrow z = \dfrac{1}{6} - \dfrac{i}{3} \\
\Rightarrow z = \dfrac{1}{6} + \left( {\dfrac{{ - 1}}{3}} \right)i{\text{ }} \to {\text{(3)}} \\
\]
By comparing equations (2) and (3), we get
Real part of the given complex part, \[x = \dfrac{1}{6}\]
Imaginary part of the given complex part, \[y = \dfrac{{ - 1}}{3}\]
As we know that the modulus of any complex number $z = x + iy$ is given by $\left| z \right| = \sqrt {{x^2} + {y^2}} {\text{ }} \to {\text{(4)}}$
Using the formula given by equation (4), we get
Modulus of the given complex number $z = \dfrac{{2 + i}}{{4i + {{\left( {1 + i} \right)}^2}}} = \dfrac{1}{6} + \left( {\dfrac{{ - 1}}{3}} \right)i{\text{ }}$ is given by
$
\Rightarrow \left| z \right| = \sqrt {{{\left( {\dfrac{1}{6}} \right)}^2} + {{\left( {\dfrac{{ - 1}}{3}} \right)}^2}} \\
\Rightarrow \left| z \right| = \sqrt {\dfrac{1}{{36}} + \dfrac{1}{9}} \\
\Rightarrow \left| z \right| = \sqrt {\dfrac{{1 + 4}}{{36}}} = \sqrt {\dfrac{5}{{36}}} \\
\Rightarrow \left| z \right| = \dfrac{{\sqrt 5 }}{6} \\
$
Since for the given complex number i.e., $z = \dfrac{{2 + i}}{{4i + {{\left( {1 + i} \right)}^2}}} = \dfrac{1}{6} + \left( {\dfrac{{ - 1}}{3}} \right)i$, the real part or x coordinate is positive and the imaginary part or y coordinate is negative. So, this complex number lies in the fourth quadrant.
Also, the principal argument (argument which lies between $ - \pi $ to $\pi $) of any complex number $z = x + iy$ is given by Principal arg(z) = $ta{n^{ - 1}}\left( {\dfrac{y}{x}} \right)$ provided this value lies between $ - \pi $ to $\pi $
Also, \[{\tan ^{ - 1}}\left( { - \theta } \right) = - {\tan ^{ - 1}}\left( \theta \right)\]
So, the principal argument of the given complex number $z = \dfrac{{2 + i}}{{4i + {{\left( {1 + i} \right)}^2}}} = \dfrac{1}{6} + \left( {\dfrac{{ - 1}}{3}} \right)i{\text{ }}$is given by
Principal arg(z) = $ta{n^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{{ - 1}}{3}} \right)}}{{\left( {\dfrac{1}{6}} \right)}}} \right] = ta{n^{ - 1}}\left( { - 2} \right) = - ta{n^{ - 1}}\left( 2 \right) = - {63.44^0}$
In order to find the general argument for any complex number lying in the fourth quadrant, we will add $2n\pi $ to its principal argument.
Argument of the given complex number is arg(z) = $2n\pi $ + Principal arg(z)
Arg(z) = $2n\pi - {63.44^0}$.
Note- In this particular problem, we have found the modulus of the given complex number by using the formula $\left| z \right| = \sqrt {{x^2} + {y^2}} $ and the argument of the given complex number obtained by $ta{n^{ - 1}}\left( {\dfrac{y}{x}} \right)$ is clearly lying between $ - \pi $ to $\pi $ or $ - {180^0}$ to ${180^0}$ so it is the principal argument. The formula for the general argument of any complex number lying in the fourth quadrant is $2n\pi + ta{n^{ - 1}}\left( {\dfrac{y}{x}} \right)$.
Complete step-by-step answer:
Let the given complex number be $z = \dfrac{{2 + i}}{{4i + {{\left( {1 + i} \right)}^2}}}{\text{ }} \to {\text{(1)}}$
As we know that the general form of any complex number is given by $z = x + iy{\text{ }} \to {\text{(2)}}$ where x is the real part of the complex number z and y is the imaginary part of the complex number z.
Now, we will convert the given complex number given by equation (1) in the same form as the general form of any complex number given by equation (2) as under
\[ \Rightarrow z = \dfrac{{2 + i}}{{4i + {1^2} + {i^2} + 2\left( 1 \right)\left( i \right)}}\]
Using ${i^2} = - 1$ in the above equation, we get
\[
\Rightarrow z = \dfrac{{2 + i}}{{4i + 1 + \left( { - 1} \right) + 2i}} \\
\Rightarrow z = \dfrac{{2 + i}}{{6i}} \\
\]
In order to rationalize the above equation, we will multiply and divide the RHS of the above equation by i, we have
i.e., \[
z = \dfrac{{i\left( {2 + i} \right)}}{{i\left( {6i} \right)}} \\
\Rightarrow z = \dfrac{{2i + {i^2}}}{{6{i^2}}} \\
\]
Using ${i^2} = - 1$ in the above equation, we get
\[
\Rightarrow z = \dfrac{{2i + \left( { - 1} \right)}}{{6\left( { - 1} \right)}} \\
\Rightarrow z = - \left( {\dfrac{{2i - 1}}{6}} \right) \\
\Rightarrow z = \dfrac{{1 - 2i}}{6} \\
\Rightarrow z = \dfrac{1}{6} - \dfrac{{2i}}{6} \\
\Rightarrow z = \dfrac{1}{6} - \dfrac{i}{3} \\
\Rightarrow z = \dfrac{1}{6} + \left( {\dfrac{{ - 1}}{3}} \right)i{\text{ }} \to {\text{(3)}} \\
\]
By comparing equations (2) and (3), we get
Real part of the given complex part, \[x = \dfrac{1}{6}\]
Imaginary part of the given complex part, \[y = \dfrac{{ - 1}}{3}\]
As we know that the modulus of any complex number $z = x + iy$ is given by $\left| z \right| = \sqrt {{x^2} + {y^2}} {\text{ }} \to {\text{(4)}}$
Using the formula given by equation (4), we get
Modulus of the given complex number $z = \dfrac{{2 + i}}{{4i + {{\left( {1 + i} \right)}^2}}} = \dfrac{1}{6} + \left( {\dfrac{{ - 1}}{3}} \right)i{\text{ }}$ is given by
$
\Rightarrow \left| z \right| = \sqrt {{{\left( {\dfrac{1}{6}} \right)}^2} + {{\left( {\dfrac{{ - 1}}{3}} \right)}^2}} \\
\Rightarrow \left| z \right| = \sqrt {\dfrac{1}{{36}} + \dfrac{1}{9}} \\
\Rightarrow \left| z \right| = \sqrt {\dfrac{{1 + 4}}{{36}}} = \sqrt {\dfrac{5}{{36}}} \\
\Rightarrow \left| z \right| = \dfrac{{\sqrt 5 }}{6} \\
$
Since for the given complex number i.e., $z = \dfrac{{2 + i}}{{4i + {{\left( {1 + i} \right)}^2}}} = \dfrac{1}{6} + \left( {\dfrac{{ - 1}}{3}} \right)i$, the real part or x coordinate is positive and the imaginary part or y coordinate is negative. So, this complex number lies in the fourth quadrant.
Also, the principal argument (argument which lies between $ - \pi $ to $\pi $) of any complex number $z = x + iy$ is given by Principal arg(z) = $ta{n^{ - 1}}\left( {\dfrac{y}{x}} \right)$ provided this value lies between $ - \pi $ to $\pi $
Also, \[{\tan ^{ - 1}}\left( { - \theta } \right) = - {\tan ^{ - 1}}\left( \theta \right)\]
So, the principal argument of the given complex number $z = \dfrac{{2 + i}}{{4i + {{\left( {1 + i} \right)}^2}}} = \dfrac{1}{6} + \left( {\dfrac{{ - 1}}{3}} \right)i{\text{ }}$is given by
Principal arg(z) = $ta{n^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{{ - 1}}{3}} \right)}}{{\left( {\dfrac{1}{6}} \right)}}} \right] = ta{n^{ - 1}}\left( { - 2} \right) = - ta{n^{ - 1}}\left( 2 \right) = - {63.44^0}$
In order to find the general argument for any complex number lying in the fourth quadrant, we will add $2n\pi $ to its principal argument.
Argument of the given complex number is arg(z) = $2n\pi $ + Principal arg(z)
Arg(z) = $2n\pi - {63.44^0}$.
Note- In this particular problem, we have found the modulus of the given complex number by using the formula $\left| z \right| = \sqrt {{x^2} + {y^2}} $ and the argument of the given complex number obtained by $ta{n^{ - 1}}\left( {\dfrac{y}{x}} \right)$ is clearly lying between $ - \pi $ to $\pi $ or $ - {180^0}$ to ${180^0}$ so it is the principal argument. The formula for the general argument of any complex number lying in the fourth quadrant is $2n\pi + ta{n^{ - 1}}\left( {\dfrac{y}{x}} \right)$.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE