Finding Values of Complex Equation

Click For Summary
SUMMARY

The discussion centers on solving the complex equation rewritten as e(i+1)(log(1-i)-log(√2)). The user successfully applies Euler's formula to express e(i+1) as e1(cos(1)+isin(1)) and derives log(1-i) as ln√2 + i(-(π/4)+2kπ). The main question raised is whether to combine these results directly or to apply the FOIL method to the exponent before proceeding. The equation ((1-i)/sqrt(2)) ^ (i+1) is also referenced as part of the solution process.

PREREQUISITES
  • Understanding of complex numbers and their properties
  • Familiarity with Euler's formula and its applications
  • Knowledge of logarithmic functions in complex analysis
  • Basic skills in algebraic manipulation, including the FOIL method
NEXT STEPS
  • Study the application of Euler's formula in complex equations
  • Learn about logarithmic identities in complex analysis
  • Research the FOIL method and its relevance in exponentiation
  • Explore the implications of the equation ((1-i)/sqrt(2)) ^ (i+1) in complex number theory
USEFUL FOR

Mathematicians, students of complex analysis, and anyone interested in solving complex equations involving logarithms and exponentials.

ver_mathstats
Messages
258
Reaction score
21
Homework Statement
Find the values of ((1-i)/sqrt(2)) ^ (i+1)
Relevant Equations
((1-i)/sqrt(2)) ^ (i+1)
I took the equation and rewrote it as: e(i+1)(log(1-i)-log(√2)).

So I worked on it in sections meaning e(i+1) and then log(1-i).

For e(i+1) I got eie1 and used Euler's formula for ei to get: e1(cos(1)+isin(1)).

And then for log(1-i) I got ln√2 + i(-(π/4)+2kπ).

Do I just bring them together now? Or would I first FOIL the exponent at the very beginning as in this part "(i+1)(log(1-i)-log(√2))"? Any help would be appreciated thank you!
 
Physics news on Phys.org
ver_mathstats said:
Relevant Equations:: ((1-i)/sqrt(2)) ^ (i+1)

I took the equation and rewrote it as: e(i+1)(log(1-i)-log(√2)).

So I worked on it in sections meaning e(i+1) and then log(1-i).
I take it
[e^{-\frac{\pi i}{4}}]^1 [e^{-\frac{\pi i}{4}}]^i
Does this work ?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K