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Babygirll41

15.05.2021

Mathematics

answer: -256

explanation

$f(f(f(f(1+$

evaluating first $f(1+i)$ , the input of $1+i$ is not real so this results in the $z^2$ piece being used:

$\nobreakspace \nobreakspace \begin{aligned}f(1+i) & = (1+i)^2 \\& = 1^2 + 2(1)(i) + i^2 & & (\text{\footnotesize using $a+b = a^2+2ab+b^2$}) \\& = 1 + 2i - 1 \\& = 2i\end{aligned}$

so we have

$\begin{aligned}f(f(f(f(1+ & = f(f(f(2i ) ) ) \end{aligned}$

since $2i$ is not real, $f(2i) = (2i)^2 = 2^2 i^2 = -4$ so

$\begin{aligned}f(f(f(f(1+ & = f(f(f(2i ) ) ) \\& = f(f(-4 ) ) \end{aligned}$

$-4$ is real, so we take the $-z^2$ piece resulting in $f(-4) = )^2 = -16$

$\begin{aligned}f(f(f(f(1+ & = f(f(-4 ) ) \\& = f(-16) \\& = )^2 \\& = -256\end{aligned}$

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