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Question: Dividing f (z) by z - I, we obtain the remainder 1 - I, and dividing it by z - I we get the remainde...

Dividing f (z) by z - I, we obtain the remainder 1 - I, and dividing it by z - I we get the remainder 1 + I, Then the remainder upon the division of f (2) by z2 +1 is
A.I + z
B.1 + z
C.1 - z
D.None of the above

Explanation

Solution

Hint : Here, z is the symbol used to denote complex number. We will use the division algorithm to get a relationship between the remainder, divisor and dividend, according to which:
f(x)=g(x)q(x)+rf(x) = g(x)q(x) + r
Where f (x) = Dividend,
g (x) = Divisor,
q (x) = Quotient
and r = remainder

** Complete step-by-step answer** :
When f (z) is divided by z - i, the remainder is i.
According to divisor algorithm:
f(z)=(z−i)q1(z)+(1−i)f(z) = (z - i){q_1}(z) + (1 - i) __________ (1)
Similarly when divided by z+i:
f(z)=(z+i)q2(z)+(1+i)f(z) = (z + i){q_2}(z) + (1 + i) ____________ (2)
Now, when f (z) is divided by (z2+1)({z^2} + 1) let the remainder be r, then:
f(z)=(z2+1)q3(z)+r(z)f(z) = ({z^2} + 1){q_3}(z) + r(z) ________ (3)
As (z2+1)({z^2} + 1) is quadratic, r will be linear and we will suppose this linear function is Az+B.
Therefore, r=Az+Br = Az + B _________ (4)
Now, equation (3) can be written as:
f(z)=(z2−i2)q3z+r(z)f(z) = ({z^2} - {i^2}){q_3}z + r(z)
[As i2 = 1]
f(z)=[(z+i)(z−i)]q3z+r(z)f(z) = \left[ {(z + i)(z - i)} \right]{q_3}z + r(z) _________ (5)
Value of z :
z−i=0⇒z=iz - i = 0 \Rightarrow z = i
z+i=0⇒z=−iz + i = 0 \Rightarrow z = - i

At z = i:
From (1) →f(i)=0+(−i) \to f(i) = 0 + ( - i)
f(i)=1−if(i) = 1 - i

From (5)→f(i)=0+r(i) \to f(i) = 0 + r(i)
f(i)=r(i)f(i) = r\left( i \right)
Therefore, r(i)=1−ir(i) = 1 - i ____ (6) (both equal to f(i)f(i))
Now, r(z) = Az +B→ri=Ai+B{\text{ r}}(z){\text{ }} = {\text{ }}Az{\text{ }} + B \to ri = Ai + B then using (6)
Ai+B=1−i.Ai + B = 1 - i. ____ (7) (both equal to r(i)r(i))

At z =-i
From (2) →f(−i)=(1+i) \to f( - i) = (1 + i)
From (5) →f(−i)=r(−i) \to f( - i) = r( - i) (as rest of the terms become 0)

Equating both :
r(−i)=1+ir( - i) = 1 + i ______ (8)
Now, r(z)=Az+B→r(−i)=−Ai+B{\text{ r}}(z) = Az + B \to r( - i) = - Ai + B then using (8)
−Ai+B=1+i- Ai + B = 1 + i _______ (9)

Adding (7) and (9):
Ai+B=1−iAi + B = 1 - i
−Ai+B=1+i- Ai + B = 1 + i
2B=22B = 2
B=1B = 1

Substituting this value of B in (7), we get:
−Ai+1=1+i- Ai + 1 = 1 + i
A=−1A = - 1

As r=Az+B,r = Az + B,its value will be
r=−z+1r = - z + 1
Or
r=1−zr = 1 - z
Therefore, when (f (z) gets divided by z2 +1, it gives 1-z as remainder and thus the correct answer is option (C).
So, the correct answer is “Option C”.

Note : The number that is to be divided is called the dividend, by which it is divided is called the divisor, the result obtained is quotient the number left at last is called the remainder.