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Richay
06-22-2006, 06:07 PM
When solving complex numbers, how do you solve it if it has negatives as the square root?

http://img512.imageshack.us/img512/6566/neg4tt.gif

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

http://img148.imageshack.us/img148/1885/neg15or.gif

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

stapel
06-22-2006, 07:25 PM
Just simplify the various square roots individually. (This is required, when working with complexes.) Then combine the "like" complex terms.

Eliz.

pka
06-22-2006, 07:36 PM
Please post the exact wording of the question.
Please do not try to interpret the meaning.

The fact is: the expression \sqrt { - 4} has no meaning in mathematics.
Many minimally prepared mathematics teachers will say the well of course that is wrong, that \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 a < 0 we do not allow the notation \sqrt a in current mathematics usage.

soroban
06-23-2006, 07:15 AM
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?
A complex number always has a negative under the square root.
\;\;That is why it is a complex number.

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

I'm guessing the problem becomes: \sqrt{-4}\,+\,\sqrt{-4}\,-\,4i^3\; . . . no
You're expected to know that: \,\sqrt{-4}\,=\,2i\, and \,\sqrt{-2}\,=\,i\sqrt{2}

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

Since \,i^2\,=\,-1\, and \,i^3\,=\,-i

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

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

And this one becomes: \,i^4\,-\,3i^2\,+\,2\sqrt{5}\;\; . . . how?
We have: \,i^4\,-\,3i^2\,+\,2(i\sqrt{2})(i\sqrt{3}) \:=\:i^4\,-\,3i^2\,+\,2i^2\sqrt{6}

Since \,i^2\,=\,-1\, and \,i^4\,=\,1

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