What are $a$ and $b$ when the zeropoints of $f(z)=(a+bi)z+2-i=0$ is at $1-i$?

Question

$f(z)=(a+bi)z+2-i$.

What are the values of a and b when $1-i$ is the zeropoint of f?

$f(z)=(a+bi)z+2-i=0$

$(a+bi)(1-i)+2-i=0$

$a+bi-ai-bi^2+2-i = 0$

$(a+b+2)+(-a+b-1)i=0$

I don't know what the next steps are.

Answer 1

So far, so good! The next step is to realize the real and imaginary parts must each be $0$, and solve the resulting linear system, which is

$a + b + 2 = 0, \tag{1}$

$-a + b - 1 = 0; \tag{2}$

it is easy to see that

$a = \dfrac{-3}{2}; \; b = \dfrac{-1}{2} \tag{3}$

is the solution.

And it's as simple as that!

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!