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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
 #1
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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
 #2
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Thanks! Your right!

Lightning  Aug 20, 2018

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