Let the modulus of (conjugate of Z) + 4i/Z equal 2 times the square root of 2. Find the modulus of Z.

Step-by-step solution

1. 1

   Choose polar form so the argument can be factored out. Write $Z = re^{i\theta}$ with $r=|Z|>0$. Then $\overline{Z} = re^{-i\theta}$ and $Z^{-1} = \tfrac1r e^{-i\theta}$ - crucially, both terms carry the same factor $e^{-i\theta}$.

2. 2

   Factor out the common exponential. $\overline{Z}+4iZ^{-1} = e^{-i\theta}\left(r+\frac{4i}{r}\right).$ Because $|e^{-i\theta}|=1$, the modulus does not depend on $\theta$ at all: $\left|\overline{Z}+4iZ^{-1}\right| = \left|r+\frac{4i}{r}\right|.$

3. 3

   Write the modulus in terms of r. The bracket has real part $r$ and imaginary part $\tfrac4r$, both real, so $\left|r+\frac{4i}{r}\right| = \sqrt{r^2+\frac{16}{r^2}} = 2\sqrt2.$

4. 4

   Square and clear the denominator. Squaring gives $r^2+\dfrac{16}{r^2} = 8$; multiplying by $r^2$ (nonzero) gives the quartic $r^4-8r^2+16 = 0.$

5. 5

   Recognise the perfect square. $r^4-8r^2+16 = (r^2-4)^2$, so $(r^2-4)^2=0$ forces $r^2=4$. There is a single repeated solution rather than two - a hint that $2\sqrt2$ is exactly the minimum possible value of the left-hand side.

6. 6

   Take the positive root and confirm. A modulus is nonnegative, so $r=|Z|=2$. Check: $\sqrt{2^2+\tfrac{16}{4}} = \sqrt{4+4} = 2\sqrt2$. By AM-GM, $r^2+\tfrac{16}{r^2} \ge 8$ with equality only at $r=2$, which is why the answer is unique.