Surprisingly big

Algebra Level 3

a and b are 2 complex values such that $a+b=1$ . Over all real values of $ab$ , what is the maximum / infimum?

1 None of the the given choices. 0.25 2 Infinity

1 solution

Kenny O.
Oct 2, 2017

This requires the following: $(x-y)(x+y)=x^2-y^2$ and $i=\sqrt{-1}$ . We are letting a be $x-y$ and b be $x+y$ ,
$(x+y)+(x-y)= 2x$ . If we make $2x=1$ , $x=0.5$ .
To maximize the use of $-y^2$ for the maximum value, $y^2$ must be the smallest negative number. This means that b can only be represented as $n*i$ where n is a real number.
If we increase the value of n, the value of ab increases while a+b is still equal to 1.

But if $b = x + y = ni$ then since $x = 0.5$ this must mean that $y = ni - 0.5$ , which in turn makes $a = x - y = 1 - ni$ .

Then $ab = (1 - ni)(ni) = ni + n^{2}$ , where $n$ is a real number. The concept of magnitude for complex numbers only makes sense if we consider the absolute value, so while asking for the maximum value of $ab$ doesn't really make sense, asking for the maximum value of $|ab|$ , i.e., the magnitude of $ab$ , would make sense and indeed give an answer of $\sqrt{n^{2} + n^{4}}$ , which goes to $\infty$ as $n \to \infty$ ..

P.S.. I made the incorrect assumption that we were only to consider real numbers so you managed to trick me. :(

Brian Charlesworth - 3 years, 8 months ago

Surprisingly big

1 solution

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