An algebra problem by Dasper Das

According to given condition, ab = +1 or ab = -1. But given condition a + b = 0 meets only when ab = -1. This is only for real numbers. But in case of complex numbers, value of ab = 1 as below:

Let us assume a = i and b = -i ( to meet a + b = 0 & a2b2 = 1)

Then ab = -(-1) = 1.

So, in question it should be mentioned.

$Given\quad that\\ \qquad a+b=0\quad ...........\quad (i)\\ \qquad { a }^{ 2 }{ b }^{ 2 }=1\quad ..........\quad (ii)\\ or\quad a=-b\quad from\quad (i)\\ Putting\quad in\quad (ii)\\ \qquad { (-b })^{ 2 }{ b }^{ 2 }=1\\ or\quad { b }^{ 4 }=1\quad =>\quad \boxed { b=1 } \\ \quad \quad \boxed { a=-1 } \qquad \qquad \qquad \qquad \because \quad a=-b\\ Now\quad find\\ \qquad \qquad ab=(-1)(1)\qquad \quad \quad \because \quad a=-1\quad \& \quad b=1\\ =>\quad \quad \quad \boxed { ab=-1 }$

$a^2 b^2 = 1$ or $(ab)^2 =1$ so $ab = 1$ or $-1$ . Now since $a+b = 0$ , $a=-b$ which means $a$ and $b$ have different parity (i.e. one positive one negative) so their product HAS to be negative so $ab = \boxed{-1}$ .

From equation (2) ab = $\pm$ 1. ${a}$ = - ${b}$ , so ab = - $b^{2}$ , therefore a = -1, as a $\leq$ 0

Let's assume that a = 1 and b = -1 and this balances the equation as 1+-1=0 and 1 1 -1 -1=1 1=1 and 1*-1=-1 which is the correct answer!!!!!!!!!!!!!

a^2 + b^2 = - 2 a b

Since this should be positive, ab has to be negative,

a^2*b^2 = 1, hence ab should be -1

We assume that a and b are reals. But with $\{a, b\}=\{i,-i\}$ then $ab=1$ so i think in the question it have noted that a,b are real numbers.