Question

Find the principle value of the argument of the complex number z=a+bi (a,b\in R, a^2+b^2\ne 0)

Collected in the board: Imaginary number

Steven Zheng posted 1 week ago

Arg (z) is the argument which lies in the interval (−π,π]. Depending on different values of a, and b, there are 8 cases.

Case 1, if a>0, b>0 when the Arg (z) lies in the first quadrant. From the definition of the argument, \tan\theta =\dfrac{b}{a}, then the principle value of the argument is the inverse tangent function, that is,

\text{Arg}(z) = \arctan\dfrac{b}{a}

Case 2, if a<0, b>0 when the point represented by the complex number lies in the second quadrant.

\text{Arg}(z) =\pi- \arctan\dfrac{b}{|a|}

Case 3, if a<0, b<0 when the point represented by the complex number lies in the quadrant III.

\text{Arg}(z) =-\pi+ \arctan\dfrac{|b|}{|a|}

Case 4, if a>0, b<0 when the \text{Arg}(z) lies in the fourth quadrant.

\text{Arg}(z) = -\arctan\dfrac{|b|}{a}

Case 5, if a=0, b>0 when the \text{Arg}(z) lies on the positive side of the imaginary axis

\text{Arg}(z)=\dfrac{\pi}{2}

Case 6, if a=0, b<0 when the \text{Arg}(z) lies on the negative side of the imaginary axis

\text{Arg}(z)=-\dfrac{\pi}{2}

Case 7 if b=0, a>0 when the \text{Arg}(z) lies on the positive side of the real axis

\text{Arg}(z)=0

Case 8 if b=0, a<0 when the \text{Arg}(z) lies on the negative side of the real axis

\text{Arg}(z)=\pi

Steven Zheng posted 1 week ago

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