Question

The complex number with the modulus $1$ and argument $\dfrac{\pi }{3}$ is denoted by $w$. Now express $w$ in the form of $x + iy$, where $x,y$ are real and exact.

Answer 1

Hint:For the given complex number $w = x + iy$ and its argument is given by ${\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$and its modulus is given by $\left| z \right| = \sqrt {{x^2} + {y^2}} $.Using these formulas we try to solve the question.

Complete step-by-step answer:
Here in the above question, we are given the complex number $w$ which is in the form of $x + iy$ and with the modulus $1$ and argument $\dfrac{\pi }{3}$.
So we can write the complex number as $z = x + iy$
$z = r(\cos \theta + i\sin \theta )$
$\cos \theta + i\sin \theta $ is also called as the $cis\theta $.
So on comparing, we get that $x = r\cos \theta ,y = r\sin \theta $
And here $r$is the modulus of the complex number which is given as $1$
So $r = 1$
So we get that $w = \cos \theta + i\sin \theta $
Argument is given by
$\arg (w) = {\tan ^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)$
And it is also given that $\arg (w) = \dfrac{\pi }{3}$
So $\dfrac{\pi }{3} = {\tan ^{ - 1}}\left( {\tan \theta } \right)$
So we know that ${\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta $
Therefore $\theta = \dfrac{\pi }{3}$
Therefore we get that
$\begin{gathered}
  x = r\cos \theta = 1\left( {\cos \dfrac{\pi }{3}} \right) = \dfrac{1}{2} \\
  y = r\sin \theta = 1\left( {\sin \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2} \\
\end{gathered} $
So our complex number is
$w = x + iy$
$ = \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i$

Note:A complex number is a number that can be expressed in the form $a + ib$, where $a$ and $b$ are real numbers, and i represents the imaginary unit.Complex number can also be written as ${e^{i\theta }}$ which is equal to the $\cos \theta + i\sin \theta $.Here i represents the iota and it is equal to $\sqrt { - 1} $ and ${i^2} = - 1$,${i^3} = - i,{i^4} = 1$.