Question on complex numbers!?

Question

When f(z) is divided by z-i the remainder is 1-i.When f(z) is divided by z+i the remainder is 1+i. What is the remainder when f(z) is divided by (z^2)+1 ?.

Please give detailed solution.

Answer 1

I think you're referring to synthetic division - maybe this wasn't clear to the previous responders.

The remainder when f(z) is divided by z^2+1 is some first degree polynomial in z, az+b:

f(z) = Q(z)(z^2+1) + (az + b)

Since z^2=(z+i)(z-i), the first term is divisible by both, and the remainders for division by (z-i) and (z+i) respectively are the same as the remainders of az+b divided by z-i and z+i:

az+b = a(z-i) + (b+ia) --> b+ia=1-i
az+b = a(z+i) + (b-ia) --> b-ia=1+i
-->
b = 1 , a = -1

In other words, the answer is:

-z + 1

Answer 2

Your supposition is in correct. Such an equation should work for all numbers but the first set that I tried(f = "6", z = "3" and i ="2") did not work. You get (f)six times (z)three (eighteen) devided by (z) three minus (i)two ... ) one. The remainder is zero, not 1 plus (i)2...
Anything drawn from an incorrect supposition will also be incorrect.

Answer 3

do you mean the quotient is real..if not so then there will be no remainder..