How do you express $\sqrt{- \frac{4}{5}}$ $sqrt(-4/5)$ as a product of a real number and i?

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How do you express $\sqrt{- \frac{4}{5}}$ $sqrt(-4/5)$ as a product of a real number and i?

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Answer 1 · 2015-12-24T21:56:25

$\sqrt{- \frac{4}{5}} = \sqrt{\frac{4}{5}} i$ $sqrt(-4/5)=sqrt(4/5)i$

Explanation:

We will be using two facts:

$i = \sqrt{- 1}$ $i = sqrt(-1)$
$\sqrt{a b} = \sqrt{a} \sqrt{b}$ $sqrt(ab) = sqrt(a) sqrt(b)$ if $a \geq 0$ $a>=0$ or $b \geq 0$ $b>=0$

With these, we have

$\sqrt{- \frac{4}{5}} = \sqrt{\frac{4}{5} \cdot (- 1)}$ $sqrt(-4/5) = sqrt(4/5*(-1))$

$= \sqrt{\frac{4}{5}} \cdot \sqrt{- 1}$ $= sqrt(4/5)*sqrt(-1)$

$= \sqrt{\frac{4}{5}} i$ $=sqrt(4/5)i$

Note that just as with non-complex square roots, if we are not speaking of the principal square root, we have

${(\sqrt{\frac{4}{5}} i)}^{2} = {(- \sqrt{\frac{4}{5}} i)}^{2} = - \frac{4}{5}$ $(sqrt(4/5)i)^2 = (-sqrt(4/5)i)^2 = -4/5$

Thus $- \frac{4}{5}$ $-4/5$ has the two possible square roots $\pm \sqrt{\frac{4}{5}} i$ $+-sqrt(4/5)i$

Answer 2 · 2015-12-25T18:31:06

$\sqrt{- \frac{4}{5}} = i \sqrt{\frac{4}{5}} = (\frac{2 \sqrt{5}}{5}) i$ $sqrt(-4/5) = i sqrt(4/5) = ((2sqrt(5))/5) i$

Explanation:

Normally speaking, every non-zero number has two square roots. In fact, in a technical sense, $0$ $0$ has two square roots, but they both happen to be $0$ $0$ .

When we use the square root symbol $\sqrt{}$ $sqrt$ we would like it to unambiguously denote one of those square roots. Then if $\sqrt{x}$ $sqrt(x)$ denotes that square root, the other is $- \sqrt{x}$ $-sqrt(x)$ .

If we can do this then the square root we denote by $\sqrt{x}$ $sqrt(x)$ is called the principal square root.

If $x > 0$ $x > 0$ then both square roots of $x$ $x$ are Real and we use the convention that $\sqrt{x}$ $sqrt(x)$ stands for the positive square root.

One of the nice bonus properties of square roots of positive numbers is that $\sqrt{a b} = \sqrt{a} \sqrt{b}$ $sqrt(ab) = sqrt(a)sqrt(b)$

If $x < 0$ $x < 0$ then both square roots of $x$ $x$ lie on the imaginary $i$ $i$ axis. By convention and definition $\sqrt{x}$ $sqrt(x)$ stands for $i \sqrt{- x}$ $i sqrt(-x)$ , that is the square root that has a positive coefficient of $i$ $i$ .

These are unambiguous and helpful definitions, leading to the answer given above.

Note however that with this definition, the property $\sqrt{a b} = \sqrt{a} \sqrt{b}$ $sqrt(ab) = sqrt(a)sqrt(b)$ no longer holds for all cases. In particular, if both $a, b < 0$ $a, b < 0$ then $\sqrt{a b} = - \sqrt{a} \sqrt{b}$ $sqrt(ab) = -sqrt(a)sqrt(b)$ . This is messy, but unavoidable if you understand what is going on when we square numbers or find square roots.

It does get more complicated when you are dealing with square roots of Complex numbers in general:

If $z = a + b i$ $z = a + bi$ with $b > 0$ $b > 0$ then its two square roots are:

$\sqrt{\frac{\sqrt{a^{2} + b^{2}} + a}{2}} + ⎛ ⎝ \sqrt{\frac{\sqrt{a^{2} + b^{2}} - a}{2}} ⎞ ⎠ i$ $sqrt((sqrt(a^2+b^2)+a)/2) + (sqrt((sqrt(a^2+b^2)-a)/2)) i$

$- \sqrt{\frac{\sqrt{a^{2} + b^{2}} + a}{2}} - ⎛ ⎝ \sqrt{\frac{\sqrt{a^{2} + b^{2}} - a}{2}} ⎞ ⎠ i$ $-sqrt((sqrt(a^2+b^2)+a)/2) - (sqrt((sqrt(a^2+b^2)-a)/2)) i$

We would probably pick the first of these as the principal square root as both its Real and imaginary parts are positive.

If $z = a + b i$ $z = a + bi$ with $b < 0$ $b < 0$ then its two square roots are:

$\sqrt{\frac{\sqrt{a^{2} + b^{2}} + a}{2}} - ⎛ ⎝ \sqrt{\frac{\sqrt{a^{2} + b^{2}} - a}{2}} ⎞ ⎠ i$ $sqrt((sqrt(a^2+b^2)+a)/2) - (sqrt((sqrt(a^2+b^2)-a)/2)) i$

$- \sqrt{\frac{\sqrt{a^{2} + b^{2}} + a}{2}} + ⎛ ⎝ \sqrt{\frac{\sqrt{a^{2} + b^{2}} - a}{2}} ⎞ ⎠ i$ $-sqrt((sqrt(a^2+b^2)+a)/2) + (sqrt((sqrt(a^2+b^2)-a)/2)) i$

In this case it is not obvious which one should be called the principal square root.

The answer depends on whether you like to consider your Complex numbers to have an angle in the range $(- π, π]$ $(-pi, pi]$ or $[0, 2 π)$ $[0, 2pi)$ .

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