Let us say I have any function $f(z)$ where $z$ denotes that this is a complex function. When we divide $f(z)$ by $z-\iota$, where $\iota= \sqrt{-1}$, we obtain a remainder $\iota$ and dividing by $z+ \iota$ we get the remainder $1+ \iota$. I want to find the remainder when $f(z)$ is divided by $z^2+1$.

Now the things clear to me here are that $$f(\iota)=\iota \tag1$$ $$f(-\iota)=1+\iota \tag2$$

Now I was thinking of taking any generalised form of $f(z)$ by which I could plug values of eqn. $(1)$ and $(2)$ into that so as to solve for the remaining variables but I am unable to deduce any general expression for $f(z)$ when it is divided by $z^2+1$ (meaning whether the resultant would be a linear expression or of higher degree). $$f(z)=z^2+1(Q(z))+R(z)$$ where $Q(z)$ and $R(z)$ are the quotients and remainders respectively. How do I ascertain any information about $Q(z)$ and $R(z)$


1 Answer 1


You can write:

$$f(z) = Q(z)(z^2 + 1) + R(z)$$

Note that $z^2 + 1 = (z+i)(z-i)$ and that $R(z)$ is (at most) a one degree polynomial in $z$ (this is what happens in general when you divide by a quadratic expression) and can be represented as $az + b$ where $a$ and $b$ are complex coefficients.


$$f(z) = Q(z)(z+i)(z-i) + az+ b$$

Also note that $f(i) = i$ and $f(-i) = 1+ i$.

From that you can get a pair of simultaneous equations:

$$i = ai + b$$

$$1+ i = -ai + b$$

and solve for $a$ and $b$. Remember that these are complex numbers, don't expect them to be purely real.

  • $\begingroup$ Should I take it as a matter of fact that $R(z)$ must surely be a one-degree polynomial? Is it always correct when we divide by a quadratic expression? $\endgroup$ Oct 5, 2016 at 0:45
  • 1
    $\begingroup$ @HarshSharma Yes. The remainder polynomial has at most (edited my answer slightly to include this subtlety) one degree less than the degree of the divisor polynomial. So when dividing by a quadratic, you get either a linear expression (degree one) or a constant (degree zero). So you can assign coefficients accordingly. $\endgroup$
    – Deepak
    Oct 5, 2016 at 0:50

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