Problem 103 - A Complete Resource Book in Mathematics for JEE Main

Question

v\(f(z)\) when divided by \(z-i\) gives remainder \(i\); when divided by \(z+i\) gives remainder \(i+1\). When \(f(z)\) is divided by \(z^{2}+1\), the remainder is (A) \(\frac{i}{2} z+\left(i-\frac{1}{2}\right)\) (B) \(\frac{i}{2} z-\left(i+\frac{1}{2}\right)\) (C) \(\frac{i}{2} z+\left(i+\frac{1}{2}\right)\) (D) \(\frac{-i}{2} z+\left(i+\frac{1}{2}\right)\)

Step-by-Step Solution

Answer

(D) \( -\frac{i}{2} z+\left(i+\frac{1}{2}\right) \)

Step 1: Understand the Problem

We are given function \( f(z) \) which leaves specific remainders when divided by linear terms \( z-i \) and \( z+i \). We need to find the remainder when \( f(z) \) is divided by the quadratic term \( z^2 + 1 \).

Step 2: Use Remainder Theorem

According to the Remainder Theorem, if \( f(z) \) is divided by \( z - a \), the remainder is \( f(a) \). Here, \( f(i) = i \) when divided by \( z-i \) and \( f(-i) = i + 1 \) when divided by \( z+i \).

Step 3: Express Remainder Using Quadratic Polynomial

Since the divisor is \( z^2 + 1 = (z-i)(z+i) \), the remainder must be a polynomial of degree less than 2, i.e., of the form \( R(z) = az + b \). We need \( R(z) \) such that \( R(i) = i \) and \( R(-i) = i+1 \).

Step 4: Solve for Remainder Coefficients

Use \( R(i) = ai + b = i \) and substitute to get the equations: \( ai + b = i \) and \( -ai + b = i + 1 \). Solve these equations:1. \( ai + b = i \)2. \(-ai + b = i + 1 \)Simplifying gives:- Add the equations to eliminate \( ai \): \( 2b = 2i + 1 \) leading to \( b = i + \frac{1}{2} \).- Subtract the first from the second equation: \( -2ai = 1 \) leading to \( ai = -\frac{1}{2} \), hence \( a = -\frac{i}{2} \).

Step 5: Construct Remainder Polynomial

Substitute \( a \) and \( b \) into \( R(z) = az + b \), obtaining the remainder polynomial as \( R(z) = -\frac{i}{2}z + \left(i + \frac{1}{2}\right) \).

Step 6: Select the Correct Answer

The remainder polynomial \( R(z) = -\frac{i}{2}z + \left(i + \frac{1}{2}\right) \) matches the option \( (D) \).

Key Concepts

Remainder Theorem

The Remainder Theorem is a useful tool in polynomial division, allowing us to easily find the remainder when dividing a polynomial by a linear factor. Let's say you have a polynomial \( f(z) \) and you want to divide it by the linear expression \( z-a \). According to the Remainder Theorem, the remainder of this division is simply \( f(a) \).
Some important points include:- If \( f(z) \) leaves a remainder when divided by \( z-a \), then \( f(a) = ext{remainder} \).- This theorem helps us quickly identify whether a given number is a root of the polynomial (if the remainder is zero).For example, in the provided exercise, dividing by \( z-i \) gives a remainder of \( i \) and dividing by \( z+i \) results in a remainder of \( i+1 \). This means that \( f(i) = i \) and \( f(-i) = i+1 \). Such information is crucial for understanding how a complex function operates mod different divisors.

Polynomial Division

Polynomial division is similar to long division with numbers, but it involves variables. It is a technique used to divide polynomials and find the quotient and the remainder. The main goal in the division of polynomials is to express one polynomial as being divisible by another, typically of lesser degree, until a remainder that cannot be further divided is obtained.
In polynomial division:- The dividend is the polynomial you are dividing into.- The divisor is the polynomial you are dividing by.- The quotient is the result of the division.- The remainder is what is left after the division.In the given exercise, we are performing division where the divisor is \( z^2 + 1 \), a quadratic polynomial. The remainder for dividing by such a polynomial will be a degree less than the divisor, typically a linear polynomial in this case.

Quadratic Polynomial

A quadratic polynomial is a polynomial of degree 2, defined in general form as \( az^2 + bz + c \). It is called quadratic because "quad" signifies square, which relates to the degree 2 term. Quadratic polynomials are fundamental to various fields of mathematics due to their simple yet profound nature.
Some properties include:

• They can have up to two roots, which may be real or complex.
• The graph of a quadratic polynomial is a parabola.

In our exercise, the divisor part of the division is \( z^2 + 1 \). We know that this expression represents a parabola that does not intersect the x-axis (in the real plane) due to its roots being \( z=i \) and \( z=-i \), which are complex roots. In this setting, dividing by \( z^2 + 1 \) entails using properties of both complex numbers and quadratic functions.

Complex Roots

Complex roots arise when solving polynomials with non-real solutions, which often occur when the discriminant of a quadratic polynomial (\( b^2 - 4ac \)) is negative. A complex number has a real part and an imaginary part and can be represented as \( a + bi \), where \( i \) is the square root of -1.
Key aspects include:

• Complex roots often appear in conjugate pairs, such as \( a + bi \) and \( a - bi \).
• They are used to express non-real roots of quadratic polynomials.

For the equation \( z^2 + 1 = 0 \) used in our exercise, the solutions are \( z = i \) and \( z = -i \), showcasing complex roots. Whereas real roots would intersect the real axis (x-axis), complex roots involve the imaginary component, which is crucial for understanding behavior in the complex plane.