Problem 8 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\tan \left[i \log \frac{a-i b}{a+i b}\right]\) is equal to (A) \(\frac{2 a b}{a^{2}+b^{2}}\) (B) \(\frac{a^{2}-b^{2}}{2 a b}\) (C) \(\frac{2 a b}{a^{2}-b^{2}}\) (D) \(a b\)

Step-by-Step Solution

Answer

(C) \(\frac{2ab}{a^2-b^2}\).

Step 1: Identify the logarithmic expression

The given expression is \(\tan \left[i \log \frac{a-i b}{a+i b}\right]\). First, we identify the logarithmic expression within the tangent function: \(i \log \frac{a-i b}{a+i b}\).

Step 2: Simplify the complex fraction

The fraction involved is \(\frac{a-i b}{a+i b}\). To simplify, consider \(z = a + ib\), which is a complex number, and its conjugate is \(\overline{z} = a - ib\). Thus, \(\frac{a-i b}{a+i b} = \frac{\overline{z}}{z}\) when expressed in terms of real and imaginary components.

Step 3: Convert the fraction to exponential form using polar coordinates

Express \(z\) in polar form as \(|z|e^{i\theta}\), where \(|z| = \sqrt{a^2 + b^2}\) and \(\theta = \arg(z) = \tan^{-1}(\frac{b}{a})\). Therefore, \(\frac{\overline{z}}{z} = \frac{|z|e^{-i\theta}}{|z|e^{i\theta}} = e^{-2i\theta}\).

Step 4: Evaluate the logarithm of the exponential

Since \(\frac{\overline{z}}{z} = e^{-2i\theta}\), we have \(\log\left(\frac{a-i b}{a+i b}\right) = \log(e^{-2i\theta}) = -2i\theta\). Substituting this back, \(\tan\left[i(-2i\theta)\right]\) becomes \(\tan(2\theta)\).

Step 5: Calculate the tangent of double angle

Use the tangent double angle formula: \(\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}\). Recall \(\tan(\theta) = \frac{b}{a}\), so \(\tan(2\theta) = \frac{2\cdot\frac{b}{a}}{1 - \left(\frac{b}{a}\right)^2} = \frac{\frac{2b}{a}}{\frac{a^2-b^2}{a^2}} = \frac{2ab}{a^2-b^2}\).

Step 6: Match with the given options

Comparing the obtained expression \(\frac{2ab}{a^2-b^2}\) to the given options, we see it matches option (C).

Key Concepts

complex logarithms

Complex logarithms extend the idea of logarithms to complex numbers. A complex number can be represented as \(z = x + yi\), where \(x\) is the real part and \(yi\) is the imaginary part. The magnitude and direction of this number in the complex plane are key to computing its logarithm.

Instead of a single value, the logarithm of a complex number yields multiple possible values, known as branches.
For example, the principal value of the logarithm of a complex number \(z\) is given by:

• \( \log z = \ln |z| + i\arg z \)

Here, \(|z|\) represents the modulus, and \(\arg z\) is its argument or angle. The formula captures both the length (modulus) and direction (argument) of the vector representing the complex number.

In our original exercise, this concept is applied to simplify the expression \(\log \left(\frac{a-ib}{a+ib}\right)\). The logarithm of the fraction of two complex numbers transforms into understanding the angles involved, by using properties from polar coordinates.

polar coordinates

Using polar coordinates can greatly simplify operations with complex numbers, such as multiplication, division, or exponentiation. In the polar form, a complex number is expressed as \(z = r e^{i\theta}\), where \(r = |z|\) is the radius or modulus, and \(\theta = \arg(z)\) is the angle with the positive real axis.

This representation converts multiplication and division of complex numbers into simpler addition and subtraction problems with the angles, while the radii are multiplied or divided.

• For a point \((a, b)\) in Cartesian coordinates, the conversion is given by:
  \(r = \sqrt{a^2 + b^2}\)
• \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\)

Thus, our step in the solution converts \(\frac{a-ib}{a+ib}\) into its exponential form using the polar equation \(e^{-2i\theta}\). The modulus in both cases cancels out. This helps simplify further calculations, like finding the complex logarithm or using trigonometric identities. Expressing complex numbers in polar coordinates makes it easy to check the effect of such transformations.

double angle formulas

The double angle formulas are a set of trigonometric identities that allow us to find the trigonometric values at double the angle. They prove particularly helpful in reducing complex expressions, such as calculating the tangent of an angle \(\theta\) doubled.

For tangent, the double angle formula is given by:

• \( \tan(2\theta) = \frac{2 \tan(\theta)}{1 - \tan^2(\theta)} \)

This formula simplifies expressions by reducing the operations needed to find \(\tan(2\theta)\).

In our problem, this formula is essential as it converts \(\tan(2\theta)\) into a simpler expression. Given \(\tan(\theta) = \frac{b}{a}\), apply the double angle formula to bridge the complex logarithmic evaluation to a straightforward trigonometric one. Calculating directly \(\tan(2\theta)\) yields \(\frac{2ab}{a^2-b^2}\). This matches the options presented, pinpointing the exact answer efficiently. Knowing the double angle formula thus helps solve problems that otherwise seem daunting.