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