Problems in understanding 2008 AMC 12B Problem 19
A function $f$ is defined by $f(z) = (4 + i) z^2 + \alpha z + \gamma$ for all complex numbers $z$, where $\alpha$ and $\gamma$ are complex numbers and $i^2 = - 1$. Suppose that $f(1)$ and $f(i)$ are both real. What is the smallest possible value of $| \alpha | + |\gamma |$ ?
Solution 2 provided by AOPS goes like this:$f(1)=4+i+\alpha+\gamma$ and$f(i)=-4-i+\alpha \cdot i +\gamma$
Since $f(1)$ and $f(i)$ are both real we get,$$\alpha+\gamma=-i$$ $$\alpha \cdot i+\gamma=i$$Solving, we get $\alpha=1-i$, $\gamma$ can be anything, to minimize the value we set $\gamma=0$, so then the answer is $\sqrt{1^2+1^2}=\sqrt{2}$. Thus, the answer is $\sqrt{2}$
However, I don't understand why we can get these equations, since for example if we just say that $\alpha+\gamma=-i$, aren't we assuming that the real part of $\alpha + \gamma = 0$?
Please let me know!
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$\begingroup$What I'd do: Consider the real and the imaginary parts of $\alpha$ and $\gamma$ separately $\alpha=\alpha_r+\alpha_i i$, $\gamma=\gamma_r+\gamma_i i $. We have then
\begin{align} f(1)&=4 +i+\alpha_r+\alpha_i i+\gamma_r+\gamma_i i\\ &= (4+\alpha_r+\gamma_r)+(1+\alpha_i+\gamma_i)i \end{align}
and
\begin{align} f(i)&=-4 -i+(\alpha_r+\alpha_i i)i+\gamma_r+\gamma_i i\\ &= (-4-\alpha_i+\gamma_r)-(1-\alpha_r-\gamma_i)i. \end{align}
In order for the function to take real values,
\begin{align} 1+\alpha_i+\gamma_i&=0\\ 1-\alpha_r-\gamma_i&=0 \end{align}
We observe right away that $\gamma_r$ can take any value and the solutions will still be real so we set $\gamma_r=0$ to minimize the value. We also observe that\begin{align} \mid\,\alpha\mid+\mid\gamma\mid&=\sqrt{\alpha_r^2+\alpha_i^2}+\sqrt{\gamma_i^2}\\ &=\{\text{using }\alpha_r=1-\gamma_i\text{ and }\alpha_i=-(1+\gamma_i)\}\\ &=\sqrt{2+2\gamma_i^2}+\sqrt{\gamma_i^2}. \end{align}This is increasing in $\gamma_i$, so we choose $\gamma_i=0$, which leaves us with $\alpha_r=1$ and $\alpha_i=-1$ and results in $\mid\,\alpha\mid+\mid\gamma\mid=\sqrt{2}$ indeed.
In fact, the solution claims that $\alpha=1-i$ (same as obtained here), i.e., the real part of $\alpha$ is not zero.
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