Complex Complexity

Algebra Level 3

(Calculate this without graphing it)

With the function:

f ( z , n ) = z 2 ⋅ n f\left(z,n\right)=z^2\cdot n

If:

c = 3 i + 6 c=3i+6

and:

f ( R e ( c ) + x , I m ( c ) + R e ( c ) ) = 4 f\left(Re\left(c\right)+x\ ,\ Im\left(c\right)+Re\left(c\right)\right)=4

What are all the possible values of x added up. Round that value to the nearest 1, then multiply it by -1.


The answer is 12.

Joshua Crawford by Joshua Crawford , 15, Australia

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1 solution

Jordan Cahn
Mar 19, 2019

R e ( c ) = 6 \mathrm{Re}(c) = 6 and I m ( c ) = 3 \mathrm{Im}(c) = 3 . So we have 4 = f ( 6 + x , 3 + 6 ) 4 = ( 6 + x ) 2 ⋅ 9 ( x + 6 ) 2 = 4 9 x + 6 = ± 2 3 x = − 6 ± 2 3 \begin{aligned} 4 &= f(6+x,3+6) \\ 4 &= (6+x)^2\cdot 9 \\ (x+6)^2 &= \frac{4}{9} \\ x+6 &= \pm \frac{2}{3} \\ x &= -6\pm\frac{2}{3} \end{aligned}

Thus x = − 16 3 x=\dfrac{-16}{3} or x = − 20 3 x=\dfrac{-20}{3} . Following the instructions at the end, we add these and multiply by − 1 -1 : − 1 ( − 16 3 − 20 3 ) = 12 -1\left(-\frac{16}{3} - \frac{20}{3}\right) = \boxed{12}

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