Find the complex number which satisfies the given equation

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Homework Help Overview

The discussion revolves around finding a complex number that satisfies a given equation involving simultaneous equations derived from the definitions of complex numbers and their conjugates.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formulation of simultaneous equations from the complex number representation, questioning the steps taken to arrive at those equations. There is a focus on verifying the correctness of the derived equations and the values of the variables.

Discussion Status

The discussion is ongoing, with participants identifying errors in calculations and expressing a need to revisit the problem. Some have offered to check their work and clarify their reasoning, while others seek further explanation of the steps involved in deriving the equations.

Contextual Notes

Participants mention constraints related to time management and the need to balance multiple assignments, which may affect their problem-solving process.

chwala
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Homework Statement
see attached
Relevant Equations
complex numbers
Find the problem here; ( i do not have the solutions...i seek alternative ways of doing the problems)

1646184142194.png
ok, i let ##z=x+iy## and ##z^*= x-iy##... i ended up with the simultaneous equation;
##2x+y=4##
##x+2y=-1##

##x=1## and ##y=2##

therefore our complex number is ##z=1+2i##
 
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chwala said:
Homework Statement:: see attached
Relevant Equations:: complex numbers

Find the problem here; ( i do not have the solutions...i seek alternative ways of doing the problems)

View attachment 297755ok, i let ##z=x+iy## and ##z^*= x-iy##... i ended up
How?
chwala said:
with the simultaneous equation;
##2x+y=4##
##x+2y=-1##

##x=1## and ##y=2##
##1+2\cdot 2 =5 \neq -1##
chwala said:
therefore our complex number is ##z=1+2i##
 
fresh_42 said:
How?

##1+2\cdot 2 =5 \neq -1##
let me check this...
 
Arrrgh:eek::oops:
##x=3## and ##y=-2## ...sign error on my simultaneous equation...i guess its best to solve this questions over the weekend...i rush on this due to other work...
 
fresh_42 said:
How?
fresh from complex numbers we know that, if ##w=a+in## then ##w^* =a-in##
 
Last edited by a moderator:
I meant: How you ended up with these equations, not how conjugation is defined, i.e. show your steps.
 
fresh_42 said:
I meant: How you ended up with these equations, not how conjugation is defined, i.e. show your steps.
Ooh ok...noted Fresh...
 

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