Answer
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Hint: A Complex number is defined as \[z = a + ib\], where ‘a’ is the real part and ‘b’ is the imaginary part.
Formula used:
Complete step-by-step answer:
Note: Argument in complex number: - It is defined as the angle inclined from the real axis in the direction of the complex number represented on the complex plane.
The argument of a complex number is defined as the angle inclined from the real axis in the direction of the complex number represented on the complex plane. It is denoted by \['\theta '\] or \['\Psi '\]. It is measured in the standard unit called ‘radians.
Formula used:
Argument of complex number formula:
In polar form, a complex number is represented by the equation \[r\left( {\cos \theta + i\sin \theta } \right)\], here, \[\theta \] is the argument. The argument function is denoted by arg(z), where z denotes the complex number i.e. \[z = x + iy\].
The computation of the complex argument can be done by using the following formula:
\[\arg \left( z \right) = \arg \left( {x + iy} \right) = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)\]
Therefore, the argument \[\theta \] is represent as:
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)\].
Let, \[z = 1 + \sqrt 2 + i\] ________ (1).
Comparing it with: \[z = x + iy\] ______ (2).
Equating the real and the imaginary part of the equation (1) & equation (2).
\[1 + \sqrt 2 + i = x + iy\]
We have,
\[\arg (z) = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)\]
\[ = {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)\]
On rationalizing:
\ = {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + \sqrt 2 }} \times \dfrac{{1 - \sqrt 2 }}{{1 - \sqrt 2 }}} \right)\ ……(using identity)
\[ = {\tan ^{ - 1}}\left( {\dfrac{{1 - \sqrt 2 }}{{{{(1)}^2} - {{(\sqrt 2 )}^2}}}} \right)\] ……\[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\]
Square of 1 is one and square root 2 is 2.
\[ = {\tan ^{ - 1}}\left( {\dfrac{{1 - \sqrt 2 }}{{1 - 2}}} \right)\]
Subtract 1 from 2 we get negative one as 1 is smaller than 2
\[ = {\tan ^{ - 1}}\left( {\dfrac{{1 - \sqrt 2 }}{{ - 1}}} \right)\]
Can also be written as
\[ = {\tan ^{ - 1}}\left( {\dfrac{{ - \left( {\sqrt 2 - 1} \right)}}{{ - 1}}} \right)\]
For this value of tan is
\[ = {\tan ^{ - 1}}\left( {\sqrt 2 - 1} \right) = \dfrac{\pi }{8}\].
\[\tan \dfrac{{{{45}^0}}}{2}\] lies in quadrant (I) and tan is positive.
Hence, \[{\tan ^{ - 1}}\left( {\sqrt 2 - 1} \right) = \dfrac{\pi }{8}\].
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