91Ó°ÊÓ

A number can be expressed in polar form r(cos θ + i sin θ) using the modulus r = \( \sqrt{x^2 + y^2} \) and the argument θ = arctan(\(\frac{y}{x}\)). Now, for the second number 2+2i, r can be calculated as \( \sqrt{2^{2} + 2^{2}} \) = \( \sqrt{8} = 2\sqrt{2} \), and θ as arctan( \( \frac{2}{2} \)) = arctan(1) = \( \frac{\pi}{4} \). Thus, the polar form is \( 2\sqrt{2} ( cos(\(\frac{\pi}{4}\)) + i sin(\(\frac{\pi}{4}\)) \). For the third number \(-\sqrt{3}+i\), r can be calculated as \( \sqrt{(-\sqrt{3})^{2} + 1^{2}} = 2 \), and θ as arctan( \( \frac{1}{-\sqrt{3}} \)) = arctan(-\( \frac{\sqrt{3}}{3} \)) = -\( \frac{\pi}{6} \). Thus, the polar form is \( 2 ( cos(-\(\frac{\pi}{6}\)) + i sin(-\(\frac{\pi}{6}\)) \).

For multiplication in polar form, multiply the magnitudes and add the angles according to the formula \( r_{1}r_{2}(cos(θ_{1}+ θ_{2})+ i sin(θ_{1}+θ_{2})). Apply this to \( i. 2\sqrt{2}.2 \) and \( (\frac{\pi}{2})+\frac{\pi}{4}-\frac{\pi}{6} \). For i, r=1, and for \( 2\sqrt{2}.2 \), r=\( 4\sqrt{2} \). The result is \( 4\sqrt{2}( cos (\frac{5\pi}{12})+ i sin(\frac{5\pi}{12})) \).

We can convert back to rectangular form using the formula \( r(cos θ + i sin θ) = r cos θ + i r sin θ. This gives \(4\sqrt{2} cos(\frac{5\pi}{12}) + i 4\sqrt{2} sin(\frac{5\pi}{12}) . After some trigonometric calculation, you find that cos(\frac{5\pi}{12}) ≈ 0.966 and sin(\frac{5\pi}{12}) ≈ 0.259. The rectangular form is approximately 3.873 + i 1.035.