I am solving a problem from a GRE guide, and stuck at the following problm.
Given that $-1 < a < 0 < \left| a \right| < b < 1$ which of the following quantity is greater?
$$\left(\frac{a^2 \sqrt{b}}{\sqrt{a}}\right)^2$$ or $$\frac{a b^5}{\left(\sqrt{b}\right)^4}$$
I don't know how to simplify first expression. Is it $\frac{a^4 b}{a}$ or $\frac{a^4 b}{-a}$
On
If you are well versed with basics of complex number, you can simply arrive at the following conclusion:
For some $a<0$, let $b=-a$. Thus $b>0$
Now $\sqrt{a} = i \sqrt{b}$ where $i = \sqrt{-1}$. Now, $$ \left(\sqrt{a}\right)^2 = \left(i\sqrt{b}\right)^2 = i^2 \left(\sqrt{b}\right)^2 = (-1) \cdot b = -b = a $$
Thus, the square of square root of a negative number is the number itself where as the square root of square of a negative number is its positive counterpart.
As an example: $$ \left(\sqrt{-2}\right)^2 = -2 \quad \text{and} \quad \sqrt{(-2)^2} = 2 $$
As written the first is undefined so they cannot be compared. They intend the first to be positive, so the denominator in your expressions should be $-a$. It does not change the answer if you use $a$ because then both are negative but the first is less so.
I believe they intend you to compare $|a^3|b$ with $ab^3$