# Question fdce1

Oct 26, 2016

$a = 1 - i$ and $b = 0$

#### Explanation:

Calling $a = {a}_{x} + i {a}_{y}$ and $b = {b}_{x} + i {b}_{y}$ we have

$f \left(z\right) = \left(4 + i\right) {z}^{2} + \left({a}_{x} + i {a}_{y}\right) z + {b}_{x} + i {b}_{y}$ so

$f \left(1\right) = 4 + {a}_{x} + {b}_{x} + i \left({a}_{y} + {b}_{y} + 1\right)$

and

$f \left(i\right) = - 4 - {a}_{y} + {b}_{x} + i \left({a}_{x} + {b}_{y} - 1\right)$

but

${a}_{y} + {b}_{y} + 1 = 0$

and

${a}_{x} + {b}_{y} - 1 = 0$

then the problem can be reduced to

$\min g \left(a , b\right) = {a}_{x}^{2} + {a}_{y}^{2} + {b}_{x}^{2} + {b}_{y}^{2}$

restricted to

${r}_{1} \left(a , b\right) = {a}_{y} + {b}_{y} + 1 = 0$
${r}_{2} \left(a , b\right) = {a}_{x} + {b}_{y} - 1 = 0$

Making the Lagrangian

$L \left(a , b , \lambda\right) = g \left(a , b\right) + {\lambda}_{1} {r}_{1} \left(a , b\right) + {\lambda}_{2} {r}_{2} \left(a , b\right)$ The stationary points are given by

{ (2 a_x + lambda_2 = 0), (2 a_y + lambda_1 = 0), (2 b_x = 0), (2 b_y + lambda_1 + lambda_2 = 0), (a_y + b_y + 1= 0), (a_x + b_y - 1= 0):}#

Solving we have

${a}_{x} = 1 , {a}_{y} = - 1 , {b}_{x} = 0 , {b}_{y} = 0 , {\lambda}_{1} = 2 , {\lambda}_{2} = - 2$

so the solution is

$a = 1 - i$ and $b = 0$

Oct 26, 2016

$| a | + | b |$ is minimized at $a = 1 - i$ and $b = 0$, giving $| a | + | b | = \sqrt{2}$.

#### Explanation:

$f \left(1\right) = \left(4 + i\right) {\left(1\right)}^{2} + a \left(1\right) + b$

$= 4 + i + a + b$

For this value to be real, the imaginary part of $a + b$ must cancel the $i$, that is, $\text{Im} \left(a + b\right) = - 1$

$f \left(i\right) = \left(4 + i\right) {\left(i\right)}^{2} + a \left(i\right) + b$

$= - 4 - i + a i + b$

For this value to be real, we must have the imaginary part of $a i + b$ cancel the $- i$, that is, $\text{Im} \left(a i + b\right) = 1$

Let $a = {a}_{r} + {a}_{i} i$ and $b = {b}_{r} + {b}_{i} i$. Then $a i = - {a}_{i} + {a}_{r} i$, so the above two observations give us the system

$\left\{\begin{matrix}{a}_{i} + {b}_{i} = - 1 \\ {a}_{r} + {b}_{i} = 1\end{matrix}\right.$

Putting ${a}_{i}$ and ${a}_{r}$ in terms of ${b}_{i}$, this gives

$\left\{\begin{matrix}{a}_{i} = - {b}_{i} - 1 \\ {a}_{r} = - {b}_{i} + 1\end{matrix}\right.$

Finally, before proceeding, notice that nowhere did we place any restrictions on ${b}_{r}$. If the above system holds, then we will have $f \left(1\right) \in \mathbb{R}$ and $f \left(i\right) \in \mathbb{R}$ regardless of the value of ${b}_{r}$. As such, we will assume ${b}_{r} = 0$ for the purpose of minimizing $| b |$.

Now, putting everything in terms of ${b}_{i}$, we have

$| a | + | b | = \sqrt{{a}_{r}^{2} + {a}_{i}^{2}} + \sqrt{{b}_{r}^{2} + {b}_{i}^{2}}$

$= \sqrt{{\left(- {b}_{i} + 1\right)}^{2} + {\left(- {b}_{i} - 1\right)}^{2}} + \sqrt{{0}^{2} + {b}_{i}^{2}}$

$= \sqrt{2 {b}_{i}^{2} + 2} + | {b}_{i} |$

By inspection, this value has a minimum at ${b}_{i} = 0$, as any change in ${b}_{i}$ in either the positive or negative direction will result in an increase in both terms. Thus, setting ${b}_{i} = 0$, we get our final answer as

$a = {a}_{r} + {a}_{i} i = \left(- {b}_{i} + 1\right) + \left(- {b}_{i} - 1\right) i = 1 - i$

$b = {b}_{r} + {b}_{i} i = 0 + 0 i = 0$

and

$| a | + | b | = \sqrt{2 {\left(0\right)}^{2} + 2} + | 0 | = \sqrt{2}$