0
$\begingroup$

I wonder if there are precise rules for squaring or making square root in both sides of an equation when we are playing with complex numbers...for example, Mathematica says this is true:

$$\left(\left(\frac{-1}{16}\right)^{\frac{1}{8}} + i*\left(\frac{-1}{16}\right)^{\frac{1}{8}}\right)^{8}=i^{2}$$

but if I make square root in both sides,then Mathematica says this is False:

$$\left(\left(\frac{-1}{16}\right)^{\frac{1}{8}} + i*\left(\frac{-1}{16}\right)^{\frac{1}{8}}\right)^{4}=i$$

We can never do that or there exist rules for doing it correctly?

$\endgroup$
4
$\begingroup$

How do you deduce from $a^8=b^2$ that $a^4=b$? This is already false for integers ($1^8=(-1)^2$, but $1^4\neq-1$) and therefore your problem is not about complex numbers at all.

The general rule is: if $a^2=b^2$, then $a=b$ or $a=-b$.

$\endgroup$
1
$\begingroup$

The assumption for taking square root is to pick the principal/positive root.

This is the reason why $\sqrt{4}=2$ excluding $-2$. However, if $x^2=4, x=2,-2$

Therefore, the procedure for your question should be

$\sqrt{\left(\left(\frac{-1}{16}\right)^{\frac{1}{8}} + i*\left(\frac{-1}{16}\right)^{\frac{1}{8}}\right)^{8}}=\sqrt{-1}$

$\left|\left(\left(\frac{-1}{16}\right)^{\frac{1}{8}} + i*\left(\frac{-1}{16}\right)^{\frac{1}{8}}\right)^{4}\right|=i$

You take the absolute value due to the assumption of taking square root (the positive one).

$\left(\left(\frac{-1}{16}\right)^{\frac{1}{8}} + i*\left(\frac{-1}{16}\right)^{\frac{1}{8}}\right)^{4}=i$ or ,$-i$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.