Complex Numbers: sqrt(-4) + sqrt(-2)sqrt(-2) - 4i^3

When solving complex numbers, how do you solve it if it has negatives as the square root?



I'm guessing the problem becomes sqrt-4 + sqrt-4 - 4i^3?



And this one becomes I^4 - 3i^2 + 2 sqrt5?

Just simplify the various square roots individually. (This is required, when working with complexes.) Then combine the "like" complex terms.

Eliz.

Please post the exact wording of the question.
Please do not try to interpret the meaning.

The fact is: the expression \(\displaystyle \sqrt { - 4}\) has no meaning in mathematics.
Many minimally prepared mathematics teachers will say the well of course that is wrong, that \(\displaystyle \sqrt { - 4} = 2i\).
But a mathematician, particularly an analyst, would accept that notation.
So I would be surprise if you were such a problem.

It is entirely proper to ask about the square root of –4!
In the complex field the answer is 2i and –2i, because i<SUP>2</SUP>=-1.

The point is if \(\displaystyle a < 0\) we do not allow the notation \(\displaystyle \sqrt a\) in current mathematics usage.

Hello, Richay!

Sorry, but I have to ask: Do you know anything about Complex Numbers?
What you're asking and what you're doing is somewhat scary . . .

When solving complex numbers, how do you solve it if it has negatives as the square root?

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A complex number always has a negative under the square root.
\(\displaystyle \;\;\)That is why it is a complex number.

\(\displaystyle \sqrt{-4}\,+\,\sqrt{-2}\sqrt{-2}\,-\,4i^3\)

I'm guessing the problem becomes: \(\displaystyle \sqrt{-4}\,+\,\sqrt{-4}\,-\,4i^3\;\) . . . no

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You're expected to know that: \(\displaystyle \,\sqrt{-4}\,=\,2i\,\) and \(\displaystyle \,\sqrt{-2}\,=\,i\sqrt{2}\)

\(\displaystyle \;\;\)The problem becomes: \(\displaystyle \,2i\,+\,(2i)(2i)\,-\,4i^3 \:=\:2i\,+\,4i^2\,-\,4i^3\)

Since \(\displaystyle \,i^2\,=\,-1\,\) and \(\displaystyle \,i^3\,=\,-i\)

\(\displaystyle \;\;\)we have: \(\displaystyle \,2i\,+\,4(-1)\,-\,3(-i)\:=\:2i\,-\,4\,+\,3i\:=\:-4\,+\,5i\)

\(\displaystyle i^4\,-\,3i^2\,+\,2\sqrt{-2)(-3)}\;\;\) . . . I hope this is a new problem!

And this one becomes: \(\displaystyle \,i^4\,-\,3i^2\,+\,2\sqrt{5}\;\;\) . . . how?

Click to expand...

We have: \(\displaystyle \,i^4\,-\,3i^2\,+\,2(i\sqrt{2})(i\sqrt{3}) \:=\:i^4\,-\,3i^2\,+\,2i^2\sqrt{6}\)

Since \(\displaystyle \,i^2\,=\,-1\,\) and \(\displaystyle \,i^4\,=\,1\)

\(\displaystyle \;\;\) we have: \(\displaystyle \,1 \,-\,3(-1)\,+\,2(-1)\sqrt{6}\:=\:1\,+\,3\,-\,2\sqrt{6} \:=\:4\,-\,2\sqrt{6}\)