Question

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

Answer 1

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.