Find the complex number which satisfies the given equation

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SUMMARY

The discussion revolves around solving a system of equations derived from a complex number representation, specifically the equations 2x + y = 4 and x + 2y = -1. The initial solution proposed was z = 1 + 2i, but upon further inspection, the correct values were identified as x = 3 and y = -2, leading to the complex number z = 3 - 2i. The conversation highlights the importance of careful algebraic manipulation and verification of solutions in complex number problems.

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  • Knowledge of complex number notation (e.g., z = x + iy)
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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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