Question

Find the modulus and argument of the complex number \[\sqrt 3 - i\] hence convert it into polar form.

Answer 1

Hint:
Complex number is a number generally represented as\[z = a + ib\], where \[a\] and \[b\] is real number represented on real axis whereas \[i\] is an imaginary unit represented on imaginary axis whose value is \[i = \sqrt { - 1} \]. Modulus of a complex number is length of line segment on real and imaginary axis generally denoted by \[\left| z \right|\] whereas angle subtended by line segment on the real axis is the argument of the matrix denoted by arg (z) calculated by trigonometric value. Argument of complex numbers is denoted by \[\arg (z) = \theta = {\tan ^{ - 1}}\dfrac{b}{a}\].

Complete step by step solution:
The given complex number in question can be written as\[z = a + ib = \sqrt 3 - i\], where \[a = \sqrt 3 \] and \[b = - 1\]
Here the modulus of complex number \[z\] will be
\[
  \left| z \right| = \sqrt {{a^2} + {b^2}} \\
   = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( { - 1} \right)}^2}} \\
   = \sqrt {3 + 1} \\
   = \sqrt 4 \\
   = 2 \\
 \]
Hence, the modulus of the complex number will be 2.
Argument of the complex:
\[
  \arg (z) = \theta \\
   = {\tan ^{ - 1}}\dfrac{b}{a} \\
   = {\tan ^{ - 1}}\dfrac{{ - 1}}{{\sqrt 3 }} \\
   = 180° - 30° \\
   = 150° \\
 \]
This can be written in radian as:
\[
  \theta = 150 \times \dfrac{\pi }{{180}} \\
   = \dfrac{{5\pi }}{6} \\
 \]

Note:
Complex numbers are always written in the form of \[z = a + ib\] where $a$ and $b$ are real numbers whereas \[i\] is an imaginary part.
We can convert a degree into radian by multiplying it by\[\dfrac{\pi }{{180}}\].