Miscellaneous Problem no 1 on logarithm of complex number

TL;DR

The video demonstrates how to solve a problem involving complex numbers and logarithms by using trigonometric ratios.

Transcript

hi friends so here in this video we are gonna see a problem based on logarithm of complex number now here the question is we have to show that tan of i log of a minus ib upon a plus ib which is a complex number is equal to 2 a b upon a square minus b square now to start with the solution i'll first of all find out the log of this complex numbers by... Read More

Key Insights

• 🎮 The video demonstrates a problem-solving technique involving complex numbers and logarithms.
• 🥳 The solution involves finding the logarithm of the complex numbers using a formula and applying trigonometric ratios.
• 🥳 The formula for logarithm of a complex number is log of the square root of the sum of the squares of the real and imaginary parts, minus i tan inverse of the ratio of the imaginary part to the real part.
• ❎ The value of i square is -1, which is used in simplifying the equation.
• 🎋 The solution is simplified by assuming a value for tan inverse of b by a and applying the formula for tan 2 theta in terms of theta.
• 🫥 The final result is the equality of tan of i log of a minus ib upon a plus ib to 2 a b upon a square minus b square.
• ❓ Inverse trigonometric functions and their properties are utilized in solving the problem.

Questions & Answers

Q: What is the problem being discussed in the video?

The problem involves showing that tan of i log of a minus ib upon a plus ib is equal to 2 a b upon a square minus b square.

Q: How is the solution obtained?

The solution involves finding the logarithm of the complex numbers using the formula for logarithm of complex numbers, subtracting the values, multiplying by i, and then finding the trigonometric ratio.

Q: What is the formula for logarithm of a complex number?

The formula for logarithm of a complex number is log of the square root of a square plus b square minus i tan inverse of b by a.

Q: How is the right-hand side of the equation simplified?

By assuming tan inverse of b by a as theta and using the formula for tan 2 theta in terms of theta, the right-hand side simplifies to 2 a b upon a square minus b square.

Summary & Key Takeaways

• The video addresses a problem related to logarithm of a complex number and aims to show that tan of i log of a minus ib upon a plus ib is equal to 2 a b upon a square minus b square.

• The solution involves finding the logarithm of the complex numbers, multiplying the answer by i, and determining the trigonometric ratio.

• By applying the formula for logarithm of complex numbers, subtracting the values, and simplifying, the required equality is proven.