# Given $f(z)=\dfrac{U(z)}{V(z)}=\dfrac{2z^3-3z^2+7z-8}{z^4-5z^3+4z^2-6z+1}$ find $f(1-\sqrt{2}i)$ without lots of complex arithmetic.

Such a problem is usually done either by direct substitution (ugh!) or synthetic division.

Synthetic division after several complex products and additions gives $U(1-\sqrt{2}i)=-8-3\sqrt{2}i$

After several more complex operations one finds, barring errors, that $V(1-\sqrt{2}i)=9+7\sqrt{2}i$ leaving the solution to the straightforward evaluation of

$$f(1-\sqrt{2}i)=\frac{-8-3\sqrt{2}i}{9+7\sqrt{2}i}$$

All standard stuff. The question is this: Can this be done without so many complex operations?

• Since the polynomials have real coefficients, you could find the remainder on dividing each of them by $z^2-2z+3$ (which has $1\pm\sqrt2 i$ as roots) to get $\frac{3x-11}{16-7x}$ and then substitute $x=1-\sqrt2 i$. But I am not sure that is any easier than the complex arithmetic. – almagest May 16 '16 at 16:07
• @almagest I found a solution based upon your suggestion to use $z^2-2z+3$. – John Wayland Bales May 17 '16 at 4:02

Since it has been 12 hours and no more answers have been suggested, I am posting a solution based upon a suggestion by @almagest.

$$[z-(1-\sqrt{2}i)]\cdot[z-(1+\sqrt{2}i)]=z^2-2z+3$$

Using long division (or better yet, quadratic synthetic division) no complex arithmetic is required to obtain

$$U(z)=(2z-1)(z^2-2z+3)+3z-11$$

and

$$V(z)=(z^2-3z-5)(z^2-2z+3)-7z+16$$

Therefore

$$U(1-\sqrt{2}i)=0+3(1-\sqrt{2}i)-11=-8-3\sqrt{2}i$$

and

$$V(1-\sqrt{2}i)=0-7(1-\sqrt{2}i)+16=9+7\sqrt{2}i$$

Therefore

\begin{align} f(1-\sqrt{2}i)&=\frac{-8-3\sqrt{2}i}{9+7\sqrt{2}i}\\ &=\frac{-8-3\sqrt{2}i}{9+7\sqrt{2}i}\cdot\frac{9-7\sqrt{2}i}{9-7\sqrt{2}i}\\ &=\frac{-114+29\sqrt{2}i}{179} \end{align}

You could try factorising it or breaking it up into simpler terms containing $z$ or $z^2$.
For example,
$U(z)$ can be broken up as
$U(z) = 2z^3 - 2z^2 + 7z - 7 - (z^2 + 1)$
$U(z) = (2z^2 + 7)(z - 1) - (z^2 + 1)$
Putting the value of z as $(1-\sqrt{2}i)$ in this expression is relatively easier.