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Question: The complex number with the modulus \(1\) and argument \(\dfrac{\pi }{3}\) is denoted by \(w\). Now ...

The complex number with the modulus 11 and argument π3\dfrac{\pi }{3} is denoted by ww. Now express ww in the form of x+iyx + iy, where x,yx,y are real and exact.

Explanation

Solution

For the given complex number w=x+iyw = x + iy and its argument is given by tan1(yx){\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)and its modulus is given by z=x2+y2\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 ww which is in the form of x+iyx + iy and with the modulus 11 and argument π3\dfrac{\pi }{3}.
So we can write the complex number as z=x+iyz = x + iy
z=r(cosθ+isinθ)z = r(\cos \theta + i\sin \theta )
cosθ+isinθ\cos \theta + i\sin \theta is also called as the cisθcis\theta .
So on comparing, we get that x=rcosθ,y=rsinθx = r\cos \theta ,y = r\sin \theta
And here rris the modulus of the complex number which is given as 11
So r=1r = 1
So we get that w=cosθ+isinθw = \cos \theta + i\sin \theta
Argument is given by
arg(w)=tan1(sinθcosθ)\arg (w) = {\tan ^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)
And it is also given that arg(w)=π3\arg (w) = \dfrac{\pi }{3}
So π3=tan1(tanθ)\dfrac{\pi }{3} = {\tan ^{ - 1}}\left( {\tan \theta } \right)
So we know that tan1(tanθ)=θ{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta
Therefore θ=π3\theta = \dfrac{\pi }{3}
Therefore we get that
x=rcosθ=1(cosπ3)=12 y=rsinθ=1(sinπ3)=32 \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+iyw = x + iy
=12+32i= \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i

Note: A complex number is a number that can be expressed in the form a+iba + ib, where aa and bb are real numbers, and i represents the imaginary unit.Complex number can also be written as eiθ{e^{i\theta }} which is equal to the cosθ+isinθ\cos \theta + i\sin \theta .Here i represents the iota and it is equal to 1\sqrt { - 1} and i2=1{i^2} = - 1,i3=i,i4=1{i^3} = - i,{i^4} = 1.