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How do you find the square root of 8i?

Answer 1

To find the square root of 8i, you can use the polar form of complex numbers. First, express 8i in polar form, then take the square root of its magnitude and halve the argument. Finally, convert the result back to rectangular form if needed.

8i in polar form is 8 * (cos(π/2) + i * sin(π/2)).

The magnitude of 8i is 8, and the argument (angle) is π/2.

The square root of 8 is 2√2.

Halving the argument gives π/4.

So, the square root of 8i is 2√2 * (cos(π/4) + i * sin(π/4)).

This can also be expressed as 2 + 2i.

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Answer 2

Nathan Axel

#2sqrt(2i)#

#sqrt(8i)#

#sqrt(4*2*i) rarr# 4 is a perfect square; it can be taken out of the radical

#2sqrt(2i)#

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Answer 3

Nathan Axel

#sqrt(8i) = 2+2i#

Take note of this:

#8i = 8(cos (pi/2) + i sin (pi/2))#

Thus, one square root according to de Moivre's theorem is:

#sqrt(8)(cos(pi/4)+isin(pi/4)) = 2sqrt(2)(1/sqrt(2)+1/sqrt(2)i) = 2+2i#

This is the principal square root. The other square root is #-2-2i#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.