charle92299
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Hi. I was wondering if someone could explain how to do the following problem. √-6 x √-15 x √-80.
Thanks!
Thanks!
Multiplying Complex Numbers: √-6 x √-15 x √-80.
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charle92299
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You don't do anything with because of the square root sign?
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Process is same ... because
√a * √b = √(a * b)
Then we have ...
√-6 x √-15 x √-80 = √[(-6) * (-15) * (-80)]
and continue.....
Subhotosh Khan and Denis, both of you got it wrong.
Let a, b, c be positive real numbers.
√(-a)*√(-b)*√(-c) =
√(-1*a)*√(-1*b)*√(-1*c) =
√(-1)*√(a)*√(-1)*√(b)*√(-1)*√(c) =
√(-1)*√(-1)*√(-1)*√(a)*√(b)*√(c) =
(i)(i)(i)*√(abc) =
-1*(i)*√(abc) =
-i√(abc) \(\displaystyle \ \ \ \ \ \ \ \ \)The correct answer has a negative sign.
- - - - - - - - - -- - - - - - - - - -- - - - - - - - - - - - - - - -
Compare it to this wrong set-up:
√(-a)*√(-b)*√(-c) = √[(-a)(-b)(-c)] = √(-abc) = √(-1*abc) = √(-1)*√(abc) = i√(abc)
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pka
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This is a flawed question even for intermediate algebra. Yes I do know it appears in many older textbooks. That does not make it correct. The trend is to never allow \(\displaystyle \sqrt{-x}\) where \(\displaystyle x>0\)(a real number). In other words, square roots of real numbers are not real numbers, we must use complex numbers.
If \(\displaystyle x<0\) then \(\displaystyle \sqrt{|x|}i\) is it principle square root in the complex number system. (Recall that \(\displaystyle i\) is defined as a number that satisfies the equation \(\displaystyle x^2+1=0\).)
Written correctly we have \(\displaystyle (\sqrt{6})(i)(\sqrt{15})(i)(\sqrt{80})(i)\) clearly gives the correct answer.
pka
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Well because \(\displaystyle -(-2)=2>0\) it is perfectly correct to write \(\displaystyle \sqrt{-(-2)}\). We are still within the real number system.
The point is, \(\displaystyle \sqrt{2}~i\) is a complex number. Yes it is very, very confusing to many who are asked to transition from the reals to the complex, do to the fact treasured rules do not carry over.
For example: \(\displaystyle \sqrt{-2}\cdot\sqrt{-8}\ne \sqrt{16}\) but is \(\displaystyle \sqrt{2}\,i\cdot\sqrt{8}\,i= -\sqrt{16}\).
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The way you have specified it,
√-6 x √-15 x √-80 = i √6 x i x √15 x i x √80 = -60 √2 i
When dealing with complex numbers and functions such as the square root, one must be careful to 'remain on the proper Riemann Sheet', see
https://people.math.osu.edu/gerlach.1/math/BVtypset/node106.html
for example. The problem with writing
√-6 x √-15 x √-80 = \(\displaystyle \sqrt{(-6) * (-15) * (-80)}\, =\, \sqrt{-7200}\)
is that one moved from one Riemann Sheet to another. More properly, in order to stay on the proper Riemann Sheet, we would have
√-6 x √-15 x √-80 = \(\displaystyle \sqrt{(6e^{i\pi}) * (15e^{i\pi}) * (80e^{i\pi})}\)
=\(\displaystyle \sqrt{7200e^{3\, \pi\, i}}\, =\, 60\, \sqrt{2}\, e^{\frac{3 \pi\, i}{2}}\) = -60 √2 i
EDIT: Oh, and as I should have continued: The problem in the other answer is basically the intermediate answer was written as
\(\displaystyle \sqrt{7200e^{3\, \pi\, i}}\, =\, \sqrt{-7200}\, =\, 60\, \sqrt{2}\, i\)
dropping a Riemann sheet [a 2\(\displaystyle \pi\) in the argument].
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There are many explanations available, such as in your textbook and online (like here). The primary point is that the rules have changed.
You've already learned about working with square roots. One of the things you learned was that sqrt(a)*sqrt(b) equalled sqrt(a*b). But you also ad the rule then that you cannot have negatives inside square roots.
Now you are allowed to have negatives inside the square roots. So clearly the rules have changed. And what you've gained in allowing negatives you then lose in multiplication. You can no longer say that sqrt(a)*sqrt(b) equals sqrt(a*b). You always first have to deal with any negatives inside the square roots. Once you pull out the imaginaries (the "i" parts), only then can you fall back on established rules.
For example:
. . . . .\(\displaystyle \sqrt{\strut -27\,}\, \cdot\, \sqrt{\strut -3\,}\, \) \(\displaystyle \color{red}{ \neq\, \sqrt{\strut 81\,}\, =\, 9}\)
Instead:
. . . . .\(\displaystyle \sqrt{\strut -27\,}\, \cdot\, \sqrt{\strut -3\,}\, =\, \sqrt{\strut -1\,}\, \sqrt{\strut 27\,}\, \cdot\, \sqrt{\strut -1\,}\, \sqrt{\strut 3\,}\, =\, i\, \sqrt{\strut 27\,}\, \cdot\, i\, \sqrt{\strut 3\,}\, \)
. . . . . . .\(\displaystyle =\, i\, \cdot\, i\, \cdot\, \sqrt{\strut 27\,}\, \cdot\, \sqrt{\strut 3\,}\, =\, i^2\, \cdot\, \sqrt{\strut 27\, \cdot 3\,} \, =\, -1\, \cdot\, \sqrt{\strut 81\,}\, =\, -9\)
Follow the same process with your exercise.
