Question

Write each complex number in polar form. Then use either (4) or (5) to obtain a polar form of the given number. Write the polar form in the form a+iba+i b. i22i\frac{-i}{2-2 i}

Solution

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Answered 2 years ago
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Let us first write z1=iz_1=-i and z2=22iz_2=2-2i in polar form before proceeding. Since z1z_1 lies on the negative part of the imaginary axis it is easy to see that its polar form is

z1=i(cos(π2)+isin(π2))=cos(π2)+isin(π2)z_1 = |i| \left( \cos \left( - \frac{\pi}{2} \right) + i \sin \left( - \frac{\pi}{2} \right) \right) =\cos \left( - \frac{\pi}{2} \right) + i \sin \left( - \frac{\pi}{2} \right)

The modulus of z2z_2 is

r2=22i=22+(2)2=4+4=22.r_2 = |2-2i| = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = 2\sqrt{2} .

Since z2z_2 lies in the fourth quadrant we know that argz2=θ\arg z_2=\theta where θ\theta is the solution to

tanθ=22=1θ=π4.\tan \theta = \frac{-2}{2} = -1 \Rightarrow \theta = - \frac{\pi}{4} .

We conclude that the polar form of z2z_2 is

z2=22(cos(π4)+isin(π4))z_2 = 2\sqrt{2} \left( \cos \left( - \frac{\pi}{4} \right) + i \sin \left( - \frac{\pi}{4} \right) \right)

Finally, we apply (5) and get

i22i=cos(π2)+isin(π2)22(cos(π4)+isin(π4))==122(cos(π2(π4))+isin(π2(π4)))==24(cos(π4)+isin(π4))==24(22isin22)=1414i\begin{aligned} \frac{-i}{2-2i} & = \frac{\cos \left( - \frac{\pi}{2} \right) + i \sin \left( - \frac{\pi}{2} \right)}{2\sqrt{2} \left( \cos \left( - \frac{\pi}{4} \right) + i \sin \left( - \frac{\pi}{4} \right) \right)} = \\ & = \frac{1}{2\sqrt{2}} \left( \cos \left( - \frac{\pi}{2} - \left( - \frac{\pi}{4} \right) \right) + i \sin \left( - \frac{\pi}{2} - \left( - \frac{\pi}{4} \right) \right) \right) = \\ & = \pmb{\frac{\sqrt{2}}{4} \left( \cos \left( - \frac{\pi}{4} \right) + i \sin \left( - \frac{\pi}{4} \right) \right)} = \\ & = \frac{\sqrt{2}}{4} \left( \frac{\sqrt{2}}{2} - i \sin \frac{\sqrt{2}}{2} \right) = \pmb{\frac{1}{4} - \frac{1}{4} i }\end{aligned}

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