The polar form of 1-√3i is [tex]4(cos 60 + isin60)[/tex]
For a complex number of the form:
z = x + iy
The polar form is of the form:
z = r (cosθ + i sinθ)
For the complex number:
1-√3i
[tex]r = \sqrt{x^2+y^2} \\r = \sqrt{1^2+(-\sqrt{3})^2 } \\r = \sqrt{1+3} \\r = \sqrt{4} \\r = 2[/tex]
[tex]\theta = tan^{-1}}\frac{-\sqrt{3} }{1} \\\theta = tan^{-1}}\sqrt{3} \\\theta = 60^0[/tex]
The polar form of the equation is therefore:
[tex]z = 4(cos 60 + isin60)[/tex]
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