Problem 33 - Advanced Engineering Mathematics

Question

Express \(\operatorname{Ln} z\) in the form \(a+i b\). \(z=(1+\sqrt{3} i)^{5}\)

Step-by-Step Solution

Answer

\( \operatorname{Ln}(z) = \ln(32) - \frac{\pi}{3} i \)

Step 1: Identify the form of complex number

Given the complex number \( z = (1+\sqrt{3}i)^5 \), we will express \( 1+\sqrt{3}i \) in polar form. A complex number can be written as \( r(\cos(\theta) + i\sin(\theta)) \), where \( r \) is the magnitude and \( \theta \) is the argument.

Step 2: Calculate the magnitude (\( r \))

The magnitude of \( 1+\sqrt{3}i \) is calculated as:\[r = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2.\]

Step 3: Determine the argument (\( \theta \))

The argument \( \theta \) of \( 1+\sqrt{3}i \) is found using the tangent function:\[\tan(\theta) = \frac{\sqrt{3}}{1} = \sqrt{3}.\]The angle whose tangent is \( \sqrt{3} \) is \( \frac{\pi}{3} \). Thus, \( \theta = \frac{\pi}{3} \).

Step 4: Express \( 1+\sqrt{3}i \) in polar form

Now that we have \( r = 2 \) and \( \theta = \frac{\pi}{3} \), we can express \( 1+\sqrt{3}i \) as:\[2 (\cos(\frac{\pi}{3}) + i \sin(\frac{\pi}{3})).\]

Step 5: Raise the polar form to the 5th power

Using De Moivre's Theorem, \((r(\cos(\theta) + i \sin(\theta)))^n\) becomes \(r^n (\cos(n\theta) + i \sin(n\theta))\). Here, with \( n = 5 \), we obtain:\[2^5 (\cos(5 \cdot \frac{\pi}{3}) + i \sin(5 \cdot \frac{\pi}{3})).\]Calculating yields:\[32 (\cos(\frac{5\pi}{3}) + i \sin(\frac{5\pi}{3})).\]

Step 6: Determine \( \cos(\frac{5\pi}{3}) \) and \( \sin(\frac{5\pi}{3}) \)

Since \( \frac{5\pi}{3} = \frac{2\pi}{3} + 2\pi = \frac{11\pi}{3} \), we adjust this within the range \(0, 2\pi\). \( \cos(\frac{5\pi}{3}) = \cos(-\frac{\pi}{3}) = \frac{1}{2} \) and \( \sin(\frac{5\pi}{3}) = \sin(-\frac{\pi}{3}) = -\frac{\sqrt{3}}{2} \).Thus, the expression becomes:\[32 \left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = 16 - 16\sqrt{3}i.\]

Step 7: Convert to logarithmic form

To convert \( z = 16 - 16\sqrt{3}i \) to logarithmic form, compute the magnitude and argument again:The magnitude is:\[|z| = \sqrt{(16)^2 + (-16\sqrt{3})^2} = \sqrt{256 + 768} = 32.\]The argument \( \phi \) using \( \tan(\phi) = \frac{-16\sqrt{3}}{16} = -\sqrt{3} \) gives \( \phi = -\frac{\pi}{3} \).

Step 8: Express \( \operatorname{Ln}(z) \) as \( a + i b \)

Finally, express the logarithm of \( z \) as:\[\operatorname{Ln}(z) = \ln(|z|) + i\phi = \ln(32) + i\left(-\frac{\pi}{3}\right).\]Therefore, \( a = \ln(32) \) and \( b = -\frac{\pi}{3} \).

Key Concepts

Polar Form

The polar form of a complex number provides insight into its magnitude and direction rather than its horizontal and vertical components. It represents complex numbers in the format \( r(\cos(\theta) + i\sin(\theta)) \), where \( r \) is the magnitude, and \( \theta \) is the angle made with the positive x-axis, known as the argument.

Magnitude \( r \) is calculated using the formula \( r = \sqrt{x^2 + y^2} \), where \( x \) and \( y \) are the real and imaginary parts of the complex number, respectively. For the number \( 1+\sqrt{3}i \), this calculation simplifies to \( r = \sqrt{1^2 + (\sqrt{3})^2} = 2 \).

The argument \( \theta \) is found using the tangent function: \( \tan(\theta) = \frac{y}{x} \). In this case, \( \theta \) becomes \( \frac{\pi}{3} \) as \( \tan(\theta) = \sqrt{3} \). By placing this in polar form, the complex number \( 1+\sqrt{3}i \) turns into \( 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})) \).

This polar representation is particularly useful when multiplying or raising complex numbers to powers, as seen with De Moivre's Theorem.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool that simplifies the process of raising complex numbers in polar form to any integer power. It's stated as \((r(\cos(\theta) + i\sin(\theta)))^n = r^n(\cos(n\theta) + i\sin(n\theta))\).

This theorem is extremely helpful because it converts the complex task of exponentiation into simple multiplications. When you want to compute \((1+\sqrt{3}i)^5\), instead of expanding it using algebraic methods, De Moivre's Theorem allows us to first convert the base to polar form, then apply the theorem.

For \( 1+\sqrt{3}i \), previously converted to \( 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})) \), raising this to the power of 5 using De Moivre’s Theorem becomes \( 2^5(\cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3})) \). After simplifying, it results in the Cartesian form, \( 16 - 16\sqrt{3}i \), illustrating how De Moivre's Theorem elegantly handles such computations.

Logarithm of a Complex Number

The logarithm of a complex number might seem less intuitive compared to real numbers, yet it opens up deeper insights in complex analysis.

Given a complex number \( z = re^{i\phi} \), its logarithm is defined as \( \ln(z) = \ln(r) + i\phi \), where \( \phi \) is the principal argument.

After determining \( z = 16 - 16\sqrt{3}i \), we calculate the magnitude \(|z| = 32 \), and the argument \( \phi = -\frac{\pi}{3} \).
Consequently, the logarithm becomes \( \operatorname{Ln}(z) = \ln(32) + i\left(-\frac{\pi}{3}\right) \).

This expression \( a + ib \), with \( a = \ln(32) \) and \( b = -\frac{\pi}{3} \), conveys both the logarithmic scale of the magnitude and the rotational effect in the complex plane. This dual nature is key for various transformations and analyses in mathematics and engineering.