Math for fun, sin(z)=2 | Summary and Q&A

TL;DR

The video explains how to find complex solutions to the equation sin(𝑧)=2 using Euler's formula and the complex logarithm.

Key Insights

• 👻 The new definition of sine in the complex world allows for complex solutions to the equation sin(𝑧)=2.
• 🥺 Euler's formula is used to rewrite sin(𝑧) in terms of 𝑒^(𝑖𝑧), leading to a quadratic equation and ultimately the solutions.
• ❓ The complex logarithm is used to find the value of ln(𝑖), which is crucial in determining the complex solutions.
• ❓ The complex solutions of sin(𝑧)=2 are 𝜋/2 + 𝑖 ln(2±√3) + 2𝜋𝑛, where 𝑛 is an integer.

Transcript

Okay, welcome to UC Berkeley, and we are at Evans Hall-- that's the math building, because right now, I'm going to do some math questions with you guys. And I'm just going to find and show you guys a classroom, and I'm going to use this classroom right here. Let's see how it is inside, and as you guys can see, the typical math classroom at UC Berke... Read More

Questions & Answers

Q: What is the new definition of sine in the complex world?

In the complex world, the new definition of sine is sin(𝑧) = (𝑒^(𝑖𝑧) - 𝑒^(-𝑖𝑧))/(2𝑖), where 𝑧 is a complex variable.

Q: How does Euler's formula relate to finding complex solutions?

Euler's formula, 𝑒^(𝑖𝑧) = cos(𝑧) + 𝑖sin(𝑧), is used to rewrite sin(𝑧) in terms of 𝑒^(𝑖𝑧), allowing for the solution of the equation sin(𝑧)=2.

Q: What is the significance of ln(𝑖) in finding the complex solutions?

ln(𝑖) is equal to 𝑖𝜋/2, which is used to calculate the complex solutions of sin(𝑧)=2 by applying the complex logarithm.

Q: Are there multiple solutions to sin(𝑧)=2?

Yes, there are multiple solutions, which are given by 𝜋/2 + 𝑖 ln(2±√3) + 2𝜋𝑛, where 𝑛 is any integer.

Summary & Key Takeaways

• The video takes place in a math classroom at UC Berkeley and introduces the concept of finding complex solutions to sin(𝑧)=2.

• The video explains the new definition of sine in the complex world using Euler's formula and the complex logarithm.

• The quadratic formula is used to solve for 𝑒^(𝑖𝑧), which leads to the solutions of the equation sin(𝑧)=2 as 𝜋/2 + 𝑖 ln(2±√3) + 2𝜋𝑛.