# How do you convert #8i# to polar form?

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To convert ( 8i ) to polar form, we need to find its magnitude (r) and argument (θ).

First, the magnitude (r) can be found using the formula: [ r = \sqrt{a^2 + b^2} ] where ( a ) is the real part and ( b ) is the imaginary part. For ( 8i ), the real part is 0 and the imaginary part is 8, so: [ r = \sqrt{0^2 + 8^2} = \sqrt{64} = 8 ]

Second, the argument (θ) can be found using the formula: [ \theta = \arctan\left(\frac{b}{a}\right) ] where ( a ) is the real part and ( b ) is the imaginary part. For ( 8i ), the real part is 0 and the imaginary part is 8, so: [ \theta = \arctan\left(\frac{8}{0}\right) ]

Since the real part is 0, we need to be careful when calculating the argument. In this case, ( \theta ) is either ( \frac{\pi}{2} ) or ( -\frac{\pi}{2} ), depending on the quadrant in which the complex number lies.

Therefore, in polar form, ( 8i ) can be expressed as: [ 8i = 8(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})) ] or [ 8i = 8(\cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2})) ]

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