Problem 15 - A Complete Resource Book in Mathematics for JEE Main 2017

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 the roots is option (B).

Step 1: Simplify the Complex Number

The given equation is \(x^{6}=(4-3i)^{5}\). Write the complex number \(4-3i\) in polar form: \(r = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5\) and \(\theta = -\tan^{-1}(3/4)\). So, \(4-3i = 5(\cos \theta + i \sin \theta)\).

Step 2: Raise the Polar Form to a Power

Using De Moivre's Theorem, \((5(\cos \theta + i \sin \theta))^5 = 5^5 (\cos(5\theta) + i \sin(5\theta))\).

Step 3: Solve for the Product of Roots

The product of all roots of the equation \(x^6 = (4-3i)^5\) is found by considering the constant term in the polynomial formed by setting \((x - \omega_1)(x - \omega_2)...(x - \omega_6) = 0\), where \(\omega_i\) are the roots. By Vieta's formulas, for an equation of the form \(x^n = a\), the product of the roots is \((-1)^n a\). Here, \(n=6\) and \(a = (4-3i)^5 = 5^5(\cos 5\theta + i \sin 5\theta)\). So, the product of the roots is \((-(5^5)(\cos 5\theta + i \sin 5\theta))\).

Step 4: Identify the Correct Option

Among the given options, \(-5^{5}(\cos 5 \theta + i \sin 5 \theta)\) is a direct match with the calculated product of all the roots from step 3.

Key Concepts

Polar Form

Complex numbers can be represented in polar form, which is very useful for multiplication, division, and raising numbers to powers. The polar form involves a modulus and an argument. To convert a complex number from its standard form, like \(a + bi\), into polar form, we calculate two things:

• The modulus \(r\), which measures distance from the origin: \(r=\sqrt{a^{2}+b^{2}}\).
• The argument \(\theta\), which is the angle formed with the positive real axis, given by \(\theta=\tan^{-1}\left(\frac{b}{a}\right)\).

Thus, the complex number can be written as \(r(\cos \theta + i \sin \theta)\). For the given problem, the polar form of \(4-3i\) is \(5(\cos \theta + i \sin \theta)\) because its modulus \(r\) is 5 and the angle \(\theta\) is derived from \(-\tan^{-1}(3/4)\). This conversion simplifies manipulation of complex numbers, especially for exponentiation.

De Moivre's Theorem

De Moivre's Theorem connects complex numbers and trigonometry in powerful ways. It states that for any complex number in polar form \(r(\cos \theta + i \sin \theta)\), raising it to the \(n\)th power can be simplified using:

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

This formula is valuable because it allows us to multiply the argument \(\theta\) and the power \(n\) to easily compute higher powers of complex numbers in their polar form. In the exercise, \((5(\cos \theta + i \sin \theta))^5 = 5^5(\cos(5\theta) + i \sin(5\theta))\) applied De Moivre's Theorem, simplifying the original problem into a straightforward expression, which is crucial to solving the exercise efficiently.

Vieta's Formulas

Vieta's formulas are powerful tools that relate the coefficients of a polynomial to sums and products of its roots. When solving polynomial equations, they can significantly simplify calculations without finding explicit root values. Particularly useful in the context of complex numbers, they assist in directly finding root properties. For instance, in a polynomial equation of degree \(n\), the product of the roots \(x^{n} = a\) is given by \((-1)^n a\).

• In this exercise, \(x^6 = (4-3i)^5\), making \(n = 6\).
• The term \(a\) is \((4-3i)^5\), calculated as \(5^5(\cos 5\theta + i \sin 5\theta)\).

Applying Vieta's formulas, the product of the roots becomes \(-5^5(\cos(5\theta) + i \sin(5\theta))\), aligning the result perfectly with one of the provided answers. This formula proves to be incredibly helpful in determining the properties of the polynomial's roots without extensive calculation.