Question

Find the value of arg\[\left( {1 + \sqrt 2 + i} \right)\].

Answer 1

Hint: A Complex number is defined as \[z = a + ib\], where ‘a’ is the real part and ‘b’ is the imaginary part.

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:

\[\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)\].

Complete step-by-step answer:

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, \[{\tan ^{ - 1}}\left( {\sqrt 2 - 1} \right) = \dfrac{\pi }{8}\].

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.