Let Z1 = 4i/(1 - i to the 555) and Z2 = (7 times i to the 88 + 24 times i to the 33)/(5i). Find the modulus of (conjugate of Z1) times Z2 squared, divided by Z1.

Step-by-step solution

1. 1

   Simplify the target with modulus rules before touching the numbers. Modulus is multiplicative, so $\left|\frac{\overline{Z_1}\,(Z_2)^2}{Z_1}\right| = \frac{|\overline{Z_1}|\,|Z_2|^2}{|Z_1|}.$ Since a conjugate has the same modulus, $|\overline{Z_1}| = |Z_1|$, and the $Z_1$ factors cancel: the answer is just $|Z_2|^2$. $Z_1$ turns out to be irrelevant - a deliberate trap in the question.

2. 2

   Reduce the powers of i using their period 4. $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, so only the exponent modulo $4$ matters. $88 \equiv 0 \pmod 4$ gives $i^{88}=1$, and $33 \equiv 1 \pmod 4$ gives $i^{33}=i$.

3. 3

   Substitute into Z2. $Z_2 = \frac{7(1)+24(i)}{5i} = \frac{7+24i}{5i}.$

4. 4

   Clear the imaginary unit from the denominator. Multiply top and bottom by $-i$, using $i\cdot(-i) = -i^2 = 1$: $Z_2 = \frac{(7+24i)(-i)}{5} = \frac{-7i-24i^2}{5} = \frac{24-7i}{5} = \frac{24}{5}-\frac{7}{5}i.$

5. 5

   Compute the squared modulus. With $a=\tfrac{24}{5}$ and $b=-\tfrac{7}{5}$: $|Z_2|^2 = a^2+b^2 = \frac{576}{25}+\frac{49}{25} = \frac{625}{25} = 25.$ The $7$-$24$-$25$ Pythagorean triple is why the answer is a clean integer.

6. 6

   Cross-check numerically. $555 \equiv 3 \pmod 4$ so $i^{555} = -i$ and $Z_1 = \tfrac{4i}{1+i} = 2+2i$. Evaluating $\left|\tfrac{(2-2i)(24/5-7i/5)^2}{2+2i}\right|$ directly gives $25.000$, confirming both the shortcut and the arithmetic.