Question

Question #fdce1

Answer 1

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

Explanation:

Calling `a=a_x+i a_y` and `b=b_x+i b_y` we have

`f(z)=(4+i)z^2+(a_x+ia_y)z+b_x+ib_y` so

`f(1)=4 + a_x + b_x + i(a_y + b_y+1)`

and

`f(i)=-4 - a_y + b_x + i(a_x + b_y-1)`

but

`a_y + b_y+1=0`

and

`a_x + b_y-1=0`

then the problem can be reduced to

`min g(a,b) = a_x^2+a_y^2+b_x^2+b_y^2`

restricted to

`r_1(a,b) = a_y + b_y+1=0`
`r_2(a,b)= a_x + b_y-1=0`

Making the Lagrangian

`L(a,b,lambda) = g(a,b)+lambda_1r_1(a,b)+lambda_2r_2(a,b)` 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`

Answer 2

`|a|+|b|` is minimized at `a = 1-i` and `b=0`, giving `|a|+|b| = sqrt(2)`.

Explanation:

`f(1) = (4+i)(1)^2 + a(1) + b`

`= 4+i + a + b`

For this value to be real, the imaginary part of `a+b` must cancel the `i`, that is, `"Im"(a+b) = -1`

`f(i) = (4+i)(i)^2 + a(i) + b`

`=-4-i+ai+b`

For this value to be real, we must have the imaginary part of `ai+b` cancel the `-i`, that is, `"Im"(ai+b) = 1`

Let `a = a_r+a_ii` and `b = b_r+b_ii`. Then `ai = -a_i+a_ri`, so the above two observations give us the system

`{(a_i+b_i = -1), (a_r+b_i = 1):}`

Putting `a_i` and `a_r` in terms of `b_i`, this gives

`{(a_i = -b_i-1), (a_r = -b_i+1):}`

Finally, before proceeding, notice that nowhere did we place any restrictions on `b_r`. If the above system holds, then we will have `f(1) in RR` and `f(i) in RR` 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((-b_i+1)^2+(-b_i-1)^2) + sqrt(0^2+b_i^2)`

`=sqrt(2b_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_ii = (-b_i+1)+(-b_i-1)i = 1-i`

`b = b_r+b_ii = 0+0i = 0`

and

`|a|+|b| = sqrt(2(0)^2+2)+|0| = sqrt(2)`