Answer on Question #58320 – Math – Complex Analysis
Question
The polar form of the complex number
z = x + i y z = x + iy z = x + i y
is given by
Solution
The algebraic form of the complex number z z z is z = x + i y z = x + iy z = x + i y .
The polar form of the complex number z z z is given by
z = A ( cos φ + sin φ ) . z = A(\cos\varphi + \sin\varphi). z = A ( cos φ + sin φ ) .
Relationship between the polar and algebraic forms of a complex number:
A = x 2 + y 2 , A = \sqrt{x^2 + y^2}, A = x 2 + y 2 , φ = { arccos x x 2 + y 2 , y ≥ 0 , x 2 + y 2 > 0 , − arccos x x 2 + y 2 , y < 0 , x 2 + y 2 > 0 , undefined , x 2 + y 2 = 0 ; \varphi = \begin{cases}
\arccos \dfrac{x}{\sqrt{x^2 + y^2}}, & y \geq 0, \sqrt{x^2 + y^2 > 0}, \\
-\arccos \dfrac{x}{\sqrt{x^2 + y^2}}, & y < 0, \sqrt{x^2 + y^2} > 0, \\
\text{undefined}, & \sqrt{x^2 + y^2} = 0;
\end{cases} φ = ⎩ ⎨ ⎧ arccos x 2 + y 2 x , − arccos x 2 + y 2 x , undefined , y ≥ 0 , x 2 + y 2 > 0 , y < 0 , x 2 + y 2 > 0 , x 2 + y 2 = 0 ;
or
φ = { arctan y x , x > 0 , + π 2 , x = 0 , y > 0 , − π 2 , x = 0 , y < 0 , arctan y x + π , x < 0 , y ≥ 0 , arctan y x − π , x < 0 , y < 0. \varphi = \begin{cases}
\arctan\dfrac{y}{x}, & x > 0, \\
+\dfrac{\pi}{2}, & x = 0, y > 0, \\
-\dfrac{\pi}{2}, & x = 0, \quad y < 0, \\
\arctan\dfrac{y}{x} + \pi, x < 0, y \geq 0, \\
\arctan\dfrac{y}{x} - \pi, x < 0, y < 0.
\end{cases} φ = ⎩ ⎨ ⎧ arctan x y , + 2 π , − 2 π , arctan x y + π , x < 0 , y ≥ 0 , arctan x y − π , x < 0 , y < 0. x > 0 , x = 0 , y > 0 , x = 0 , y < 0 ,
Relationship between the algebraic and polar forms of a complex number:
x = A cos φ , y = A sin φ . x = A\cos\varphi, y = A\sin\varphi. x = A cos φ , y = A sin φ .
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