Question

Find the principal argument of the complex number \[1 + \sqrt 2 + i\]. Choose the correct option.
A. \[\dfrac{\pi }{3}\]
B. \[\dfrac{\pi }{6}\]
C. \[\dfrac{\pi }{8}\]
D. \[\dfrac{\pi }{4}\]

Answer 1

Hint: The complex number \[z\] is of the form \[z = x + iy\] where \[x\] represents the real part and \[y\] represents the imaginary part. In the given question, \[z = 1 + \sqrt 2 + i\] has both real and imaginary parts that is \[x = 1 + \sqrt 2 \] is the real part of the complex number and \[y = 1\] is the imaginary part of the given complex number. Use the formula of finding the principal argument of the complex number to find the argument of \[z\]. Use the Pythagoras theorem to find the value of \[\theta \].

Complete step by step solution:
Consider the given complex number \[z = x + iy\] which has both the parts in it. The real part is \[x = 1 + \sqrt 2 \] and the imaginary part is \[y = 1\].
We know that, for a complex number \[z = x + iy\], the principal argument of the complex number is given by \[\arg \left( z \right) = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)\].
Next, we will substitute the values in the formula of principal argument of \[z\].
Thus, we get,
\[\arg \left( {1 + \sqrt 2 + i} \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)\]
Now, we will find the value of \[{\tan ^{ - 1}}\left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)\] by multiplying the numerator and denominator by \[1 - \sqrt 2 \]
Thus, we get,
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + \sqrt 2 }} \times \dfrac{{1 - \sqrt 2 }}{{1 - \sqrt 2 }}} \right)\]
Next, we will use the property \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\] in the denominator.
Hence, we get,
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{1 - \sqrt 2 }}{{1 - 2}}} \right) = {\tan ^{ - 1}}\left( {\sqrt 2 - 1} \right)\]
Now, we will find the value of \[{\tan ^{ - 1}}\left( {\sqrt 2 - 1} \right) = \theta \]
\[ \Rightarrow \tan \left( \theta \right) = \sqrt 2 - 1\]
Consider the figure,


From this we have, \[AB = \sqrt 2 - 1\] and \[BC = 1\] and we have to find the value of \[AC = ?\].
Using the Pythagoras theorem, we know that for a \[\Delta ABC\], \[{\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2}\]
Thus, we get,
\[
   \Rightarrow {\left( {AC} \right)^2} = {\left( {\sqrt 2 - 1} \right)^2} + {\left( 1 \right)^2} \\
   \Rightarrow {\left( {AC} \right)^2} = 2 + 1 - 2\sqrt 2 + 1 \\
   \Rightarrow {\left( {AC} \right)^2} = 4 - 2\sqrt 2 \\
   \Rightarrow \left( {AC} \right) = \sqrt {4 - 2\sqrt 2 } \\
   \Rightarrow AC = \sqrt {4\left( {1 - \dfrac{{2\sqrt 2 }}{4}} \right)} \\
   \Rightarrow AC = 2\sqrt {1 - \dfrac{{\sqrt 2 }}{2}} \\
   \Rightarrow AC = 2\sqrt {1 - \dfrac{1}{{\sqrt 2 }}} \\
   \Rightarrow AC = 2\sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}} \\
\]
Now, we have \[AB = \sqrt 2 - 1\], \[BC = 1\] and \[AC = 2\sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}} \]
Next, we will find the value of \[\sin \theta = \dfrac{{AB}}{{AC}}\] and \[\cos \theta = \dfrac{{BC}}{{AC}}\]
Thus, we have,
\[ \Rightarrow \sin \theta = \dfrac{{\sqrt 2 - 1}}{{2\sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}} }}\] and \[\cos \theta = \dfrac{1}{{2\sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}} }}\]
Next, we will evaluate the value of \[\sin \theta \cos \theta \]
\[
  \sin \theta \cos \theta = \dfrac{{\sqrt 2 - 1}}{{2\sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}} }} \times \dfrac{1}{{2\sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}} }} \\
   = \dfrac{{\sqrt 2 - 1}}{{4\left( {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}} \right)}} \\
   = \dfrac{{\sqrt 2 }}{4} \\
   = \dfrac{1}{{2\sqrt 2 }} \\
\]
Now, multiply 2 on both the sides, we have,
\[
  2\sin \theta \cos \theta = \dfrac{2}{{2\sqrt 2 }} \\
  \sin 2\theta = \dfrac{1}{{\sqrt 2 }} \\
  2\theta = \dfrac{\pi }{4} \\
  \theta = \dfrac{\pi }{8} \\
 \]
Thus, the principal argument \[\arg \left( {1 + \sqrt 2 + i} \right) = \dfrac{\pi }{8}\].
The option is incorrect as the principal argument is equal to \[\dfrac{\pi }{8}\].
Note: The trigonometric identity \[2\sin \theta \cos \theta = \sin 2\theta \] is used to simplify the expression. Remember the trigonometric values as here we have used
\[ \sin A = \dfrac{1}{{\sqrt 2 }} \] giving \[A = \dfrac{\pi }{4} \]. We have used the Pythagoras theorem in evaluating the value of \[\theta \] by finding the value of the side of the right-angled triangle.