Respuesta :

Answer:

[tex]-\sqrt{2} + \sqrt{2}i[/tex]

Step-by-step explanation:

Angle of 9pi/4

The equivalent angle of [tex]\frac{9\pi}{4}[/tex], on the first lap, is found subtracting this angle from [tex]2\pi[/tex]. Thus:

[tex]\frac{9\pi}{4} - 2\pi = \frac{9\pi}{4} - \frac{8\pi}{4} = \frac{\pi}{4}[/tex]

Thus, the sine and cosine are:

[tex]\sin{(\frac{9\pi}{4})} = \sin{(\frac{\pi}{4})} = \frac{\sqrt{2}}{2}[/tex]

[tex]\cos{(\frac{9\pi}{4})} = \cos{(\frac{\pi}{4})} = \frac{\sqrt{2}}{2}[/tex]

Angle of 3pi/2

On the first lap of the circle, thus no need to find the equivalent angle. We have that:

[tex]\sin{(\frac{3\pi}{2})} = -1, \cos{(\frac{3\pi}{2})} = 0[/tex]

Expression:

[tex]4(\cos{(\frac{9\pi}{4})} + i\sin{(\frac{9\pi}{4})}) \div 2(\cos{(\frac{3\pi}{2})} + i\sin{(\frac{3\pi}{2})})[/tex]

[tex]4(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}) \div 2(0 - i)[/tex]

[tex]\frac{2\sqrt{2} + 2\sqrt{2}i}{-2i} \times \frac{i}{i}[/tex]

Considering that [tex]i^2 = -1[/tex]

[tex]\frac{-2\sqrt{2}+2\sqrt{2}i}{2}[/tex]

[tex]-\sqrt{2} + \sqrt{2}i[/tex]

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