Complex Set

Writing $z$ in the polar form: $z=cis\alpha$ where $\alpha$ is the argument of the complex number:

Then $a=|cis2\alpha +cis\alpha -1|$ $\Longleftrightarrow$ $a=|cis2\alpha +cis\alpha -cis0|$ $\Longleftrightarrow$ $a=|(cis\alpha)(cis\alpha +1 -cis(-\alpha))|$

But $cis\alpha + cis(-\alpha)$ equals to $z - \bar { z }$ that equals to $2i sin\alpha$

Then $a$ equals to $|(cis\alpha)(1+2isin\alpha)|$

Property: $|z.w|=|z|.|w|$ hence, $a=|cis\alpha||1+2isin\alpha|$ $\Longleftrightarrow$ $a=\sqrt { { 1 }^{ 2 }+4\sin ^{ 2 }{ \alpha } }$ which is maximized when $sin\alpha$ equals to $1$ .

So, the maximum value of $a$ is $\boxed { \sqrt { 5 } }$

Let's try a calculus solution to this problem. Let $z = x+yi$ such that $x^2+y^2 =1$ . If $a = |z^2+z-1|$ , then:

$a = |(x+yi)^2 + (x+yi) - 1| = \sqrt{(x^2-y^2+x-1)^2 + (2xy + y)^2};$

or $a = \sqrt{[x^2+x-1 - (1-x^2)]^2 + y^2(2x+1)^2};$

or $a = \sqrt{(2x^2+x -2)^2 + (1-x^2)(2x+1)^2};$

or $a = f(x) = \sqrt{5-4x^2}.$

Taking the first derivative equal to zero yields: $f'(x) = -\frac{4x}{\sqrt{5-4x^2}} = 0 \Rightarrow x = 0$ . The second derivative at $x = 0$ produces:

$f''(x) = -\frac{20}{(5-4x^2)^{3/2}} \Rightarrow f''(0) = -\frac{4}{\sqrt{5}} < 0$

hence a global maximum. Finally, $a_{MAX} = \sqrt{5-4 \cdot 0^2} = \boxed{\sqrt{5}}.$