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Question: The \[\arg \left( { - \dfrac{3}{2}} \right)\] equals A. \[\dfrac{\pi }{2}\] B. \[ - \dfrac{\pi ...

The arg(32)\arg \left( { - \dfrac{3}{2}} \right) equals
A. π2\dfrac{\pi }{2}
B. π2 - \dfrac{\pi }{2}
C.0
D. π\pi

Explanation

Solution

Hint : Here in this question, we have to find the angle of the complex number using a given argument number. As we know the complex number is defined as z=x+iyz = x + iy , where x=rcosθx = r\cos \theta , y=rsinθy = r\sin \theta and ii be the imaginary number by giving the value of r to the polar form of complex number z=r(cosθ+isinθ)z = r\left( {\cos \theta + i\sin \theta } \right) using a given argument number we get the angle θ\theta .

Complete step-by-step answer :
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 “ θ\theta ”. It is measured in the standard unit called “radians”.
In polar form, a complex number is represented by the equation z=r(cosθ+isinθ)z = r\left( {\cos \theta + i\sin \theta } \right) , here, θ\theta is the argument. The argument function is denoted by arg(z)\arg \left( z \right) , where z denotes the complex number, i.e., z=x+iyz = x + iy . The computation of the complex argument can be done by using the following formula:
i.e., arg(z)=θ\arg \left( z \right) = \theta
Therefore, the argument θ is represented as: θ=tan1(yx)\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)
Now, consider the given question
arg(z)=arg(32)\Rightarrow \,\,\arg \left( z \right) = \arg \left( { - \dfrac{3}{2}} \right)
by
Where, z is the complex number i.e., z=x+iyz = x + iy , then
arg(x+iy)=arg(32)\Rightarrow \,\,\arg \left( {x + iy} \right) = \arg \left( { - \dfrac{3}{2}} \right)
Let us take
x+iy=(32)\Rightarrow \,\,x + iy = \left( { - \dfrac{3}{2}} \right)
Put, x=rcosθx = r\cos \theta and y=rsinθy = r\sin \theta , then on substituting we have
rcosθ+irsinθ=(32)\Rightarrow \,\,r\cos \theta + i\,r\sin \theta = \left( { - \dfrac{3}{2}} \right)
Take r as common in LHS, then
r(cosθ+isinθ)=(32)\Rightarrow \,\,r\left( {\cos \theta + i\,\sin \theta } \right) = \left( { - \dfrac{3}{2}} \right)
Now, put r=32r = \dfrac{3}{2} and θ=π\theta = \pi , then
32(cos(π)+isin(π))=(32)\Rightarrow \,\,\dfrac{3}{2}\left( {\cos \left( \pi \right) + i\,\sin \left( \pi \right)} \right) = \left( { - \dfrac{3}{2}} \right)
By the standard trigonometric table the value of cos(π)=1\cos \left( \pi \right) = - 1 and sin(π)=0\sin \left( \pi \right) = 0 , on substituting the values we have
32(1+i(0))=(32)\Rightarrow \,\,\dfrac{3}{2}\left( { - 1 + i\,\left( 0 \right)} \right) = \left( { - \dfrac{3}{2}} \right)
32(1)=(32)\Rightarrow \,\,\dfrac{3}{2}\left( { - 1} \right) = \left( { - \dfrac{3}{2}} \right)
32=32\Rightarrow \,\, - \dfrac{3}{2} = - \dfrac{3}{2}
Hence, arg(32)=π\arg \left( { - \dfrac{3}{2}} \right) = \pi
Therefore, option (D) is correct.
So, the correct answer is “Option D”.

Note : A complex number are one of the numbers that are expressed in the form of a+iba + ib , where a,b be the real number and ii be an imaginary number, absolute number is an angle towards the direction of the complex number it can easily find by a formula of θ=tan1(yx)\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) , where, y=rsinθy = r\sin \theta and x=rsinθx = r\sin \theta .