The Cartesian form of the complex number z = | vedclass

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Question

(A) Given the complex number in polar form: $z = 4(\cos 300^{\circ} + i \sin 300^{\circ})$.We know that $\cos 300^{\circ} = \cos(360^{\circ} - 60^{\ci

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$2 - 2\sqrt{3}i$

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(A) Given the complex number in polar form: $z = 4(\cos 300^{\circ} + i \sin 300^{\circ})$.We know that $\cos 300^{\circ} = \cos(360^{\circ} - 60^{\circ}) = \cos 60^{\circ} = \frac{1}{2}$.And $\sin 300^{\circ} = \sin(360^{\circ} - 60^{\circ}) = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2}$.Substituting these values into the expression:$z = 4\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}ight)$.$z = 4 	imes \frac{1}{2} - 4 	imes i\frac{\sqrt{3}}{2}$.$z = 2 - 2\sqrt{3}i$.

The Cartesian form of the complex number $z = 4(\cos 300^{\circ} + i \sin 300^{\circ})$ is

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