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# Piecewise-Defined Functions

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For all complex numbers z, let $$f(z) = \left\{ \begin{array}{cl} z^{2}&\text{ if }z\text{ is not real}, \\ -z^2 &\text{ if }z\text{ is real}. \end{array} \right.$$

Find $$f(f(f(f(1+i))))$$.

Aug 20, 2018

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Let's define $$z_0=1+i$$. Then we want to find $$f(f(f(f(z_0))))$$.

z is not real, so we have$$f(z_0)=z_0^2=(1+i)^2=1+2i+i^2=1+2i-1=2i$$.

So $$f(z_0)=2i$$ is not real either, and therefore $$f(f(z_0))=f(z_0)^2=(2i)^2=-4.$$

Now things change, because $$f(f(z_0))=-4$$ is real and so $$f(f(f(z_0)))=-(-4)^2=-16$$.

This is still real, so we have finally:

$$f(f(f(f(z_0))))=-(-16)^2=-256$$

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Aug 20, 2018
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