Problem 15 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(x^{6}=(4-3 i)^{5}\), then the product of all of its roots is (where \(\left.\theta=-\tan ^{-1}(3 / 4)\right)\) (A) \(5^{5}(\cos 5 \theta+i \sin 5 \theta)\) (B) \(-5^{5}(\cos 5 \theta+i \sin 5 \theta)\) (C) \(5^{5}(\cos 5 \theta-i \sin 5 \theta)\) (D) \(-5^{5}(\cos 5 \theta-i \sin 5 \theta)\)

Step-by-Step Solution

Answer

The product of all roots is option (B): \(-5^5(\cos 5 \theta+i \sin 5 \theta)\).

Step 1: Express as Polar Form

First, convert the complex number \(4 - 3i\) into its polar form. The modulus \(r\) is given by \(r = \sqrt{4^2 + (-3)^2} = 5\). The argument \(\theta\) is given by \(\theta = -\tan^{-1}(3/4)\). Thus, the polar form is \(5 (\cos \theta + i \sin \theta)\).

Step 2: Raise Polar Form to Power of 5

Raise \( (5(\cos \theta + i \sin \theta)) \) to the power of 5 using De Moivre's Theorem: \( (r (\cos \theta + i \sin \theta))^5 = r^5 (\cos(5\theta) + i \sin(5\theta)) \). Thus, \( (4 - 3i)^5 = 5^5 (\cos(5\theta) + i \sin(5\theta)) \).

Step 3: Apply Root Properties

The equation \(x^6 = (4 - 3i)^5\) implies that \(x^6 = 5^5 (\cos(5\theta) + i \sin(5\theta))\). The 6th roots of this complex number will also have modulus \(5^{5/6}\).

Step 4: Find Product of Roots

The product of all the roots of the equation \(x^6 = (4 - 3i)^5\) is the product of a complete set of 6th roots of \(5^5 (\cos(5\theta) + i\sin(5\theta))\). By the properties of complex numbers, the product of all \(n\)th roots of a complex number is \(- \text{modulus of roots}\), thus \(- (5^{5/6})^6 = -5^5\).

Step 5: Identify Correct Answer

Thus, the expression for the product of all the roots considering the angle \(5\theta\) is \(-5^5 (\cos(5\theta) + i\sin(5\theta))\). Therefore, the correct option is (B).

Key Concepts

Polar Form

The polar form is a powerful way to express complex numbers. It represents a complex number in terms of its magnitude, known as the modulus, and its angle, known as the argument. When you have a complex number such as \(4 - 3i\), the polar form transforms it into \(r (\cos(\theta) + i \sin(\theta))\).

Finding the modulus \(r\) is straightforward. It is calculated using the Pythagorean theorem, \(r = \sqrt{a^2 + b^2}\), where \(a\) and \(b\) are the real and imaginary parts, respectively. In our example with \(4 - 3i\), \(r\) equals 5.

The angle \(\theta\) is found using the arctangent function, \(\theta = \tan^{-1}(b/a)\). Here, it results in \(\theta = -\tan^{-1}(3/4)\). So, the complex number \(4 - 3i\) in polar form is \(5(\cos(\theta) + i \sin(\theta))\).

This form is particularly useful when performing multiplication or raising numbers to a power.

De Moivre's Theorem

De Moivre's Theorem is a handy tool in complex number arithmetic, especially when raising complex numbers in polar form to any power. The theorem states that for any complex number \(r(\cos(\theta) + i \sin(\theta))\) and integer \(n\), the expression raised to the power of \(n\) is given as:

• \[r^n (\cos(n\theta) + i \sin(n\theta))\]

For example, to calculate \((4 - 3i)^5\), first convert it to polar form as \(5(\cos(\theta) + i \sin(\theta))\). Applying De Moivre's Theorem, elevate the modulus to the fifth power: \[5^5(\cos(5\theta) + i \sin(5\theta))\].

This method vastly simplifies operations involving powers, converting complex multiplications into simple arithmetic.

Roots of Equations

Finding the roots of a complex equation involves determining all possible solutions that satisfy it. For an equation like \(x^6 = (4 - 3i)^5\), six solutions (or roots) can be identified.

Using the polar form simplifies finding these roots. Each root is equally spread around a circle in the complex plane. If \(z = r(\cos(\theta) + i \sin(\theta))\) is a solution, then the \(n\)th roots will all have the same modulus \(r^{1/n}\) and angles distributed as \(\theta_k = \frac{\theta + 2k\pi}{n}\), with \(k = 0, 1, \, ..., n-1\).

In our specific problem, the sixth root of \(5^5(\cos(5\theta) + i \sin(5\theta))\) would be calculated with a modulus of \(5^{5/6}\), with the angles incrementing by \(\frac{2\pi}{6}\).

Complex Number Modulus

The modulus of a complex number is akin to its absolute value in the real number system. It measures the "size" or "distance" of the complex number from the origin in the complex plane. For a complex number \(a + bi\), the modulus \(r\) is computed with the formula \(r = \sqrt{a^2 + b^2}\).

Modulus matters greatly in operations such as determining powers and roots of complex numbers. For example, when raising a complex number to a power or extracting roots, the modulus itself is raised to the power or root. In our exercise, \((4 - 3i)^5\) results in a modulus of \(5^5\), and those roots share a modulus of \(5^{5/6}\).

The modulus is also pivotal when considering De Moivre's Theorem as it influences the resulting magnitude of the complex number after exponentiation.