Complex Numbers: simplify [ ( 1 / 2i )^2 ] * ( -2i )^2

[charle92299]

Can someone answer and explain how to do the following problem: [(1/2i)²] * (-2i)²

 

[Deleted member 4993]

Can someone answer and explain how to do the following problem: [(1/2i)²] * (-2i)²

How much is:

i² = ?

(-i)² = ?

What are your thoughts?

Please share your work with us ...even if you know it is wrong

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[lookagain]

Can someone answer and explain how to do the following problem: [(1/2i)²] * (-2i)²

Did you intend \(\displaystyle \ \bigg[\bigg(\dfrac{1}{2}i\bigg)^2\bigg](-2i)^2 \ ?\)


Or, did you intend \(\displaystyle \ \bigg[\bigg(\dfrac{1}{2i}\bigg)^2\bigg](-2i)^2 \ ?\)

 

[pka]

Did you intend \(\displaystyle \ \bigg[(\dfrac{1}{2}i\bigg)^2\bigg](-2i)^2 \ ?\)

Or, did you intend \(\displaystyle \ \bigg[\bigg(\dfrac{1}{2i}\bigg)^2\bigg](-2i)^2 \ ?\)

Thank you for pointing out the need for grouping symbols. We should all stress their use in every posting.

The irony is in this case it makes makes no difference: \(\displaystyle \bigg[(\dfrac{1}{2}i\bigg)^2\bigg](-2i)^2 =\ \bigg[\bigg(\dfrac{1}{2i}\bigg)^2\bigg](-2i)^2 \ \)
Reason: \(\displaystyle i^n=\pm 1\), a real number for any even integer \(\displaystyle n\).