How do you find the roots of the complex equation cos(z)=2?

Click For Summary
SUMMARY

The discussion focuses on finding the roots of the complex equation cos(z) = 2, where z is a complex number. The solution involves expressing cos(z) using the exponential function derived from Euler's formula: cos(z) = (e^(iz) + e^(-iz))/2. By substituting z = a + bi into this equation, the transformed equation e^(iz) + e^(-iz) = 4 can be solved to find the roots.

PREREQUISITES
  • Understanding of complex numbers and their representation (z = a + bi).
  • Familiarity with Euler's formula and its application in trigonometric functions.
  • Knowledge of exponential functions and their properties.
  • Basic skills in solving equations involving complex variables.
NEXT STEPS
  • Study the derivation and application of Euler's formula in complex analysis.
  • Learn how to manipulate and solve equations involving complex exponentials.
  • Explore the implications of trigonometric functions in the complex plane.
  • Research methods for finding roots of complex equations, including graphical and numerical techniques.
USEFUL FOR

Students studying complex analysis, mathematicians interested in trigonometric equations, and anyone seeking to deepen their understanding of complex functions and their roots.

buzzmath
Messages
108
Reaction score
0

Homework Statement


Find all roots of the equation cos(z)=2 (z is a complex number)

Homework Equations


The Attempt at a Solution


What do they mean find the roots of this equation? We're just going over trig functions and it doesn't say anything about roots so I'm not sure what they're asking for. when I looked on the internet I just kept getting things raised to 1/n. I don't see that all here.
 
Last edited:
Physics news on Phys.org
Okily well the roots basically means the solutions. We want a z that which, cos z=2 ok?

Write out cos x in terms of the exponential function, (which can be derived from Euler Formula).

cos x = \frac{e^{ix} + e^{-ix}}{2}. Now write z in terms of its reals and imaginarys, z=a + bi. Substitute and get

e^{iz} + e^{-iz} = 4. Hopefully you can work it from there.
 

Similar threads

  • · Replies 22 ·
Replies
22
Views
1K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K