Solve this pair of simultaneous equations involving complex numbers

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Homework Help Overview

The discussion revolves around solving a pair of simultaneous equations involving complex numbers. The original poster presents two equations and attempts to manipulate them to find the values of the variables involved.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore different methods for solving the equations, including direct manipulation and matrix representation. Some question the efficiency of the approaches taken, while others suggest alternative methods like Cramer's Rule.

Discussion Status

The discussion includes various attempts to solve the equations, with some participants providing insights into their methods. There is no explicit consensus on the best approach, but multiple interpretations and techniques are being explored.

Contextual Notes

Participants discuss formatting issues related to LaTeX typesetting, which may affect the presentation of mathematical expressions. There is also mention of the potential for different approaches to yield similar results.

chwala
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Homework Statement
Solve the simultaneous equation for the complex number ##z## and ##w##,

$$(1+i)z+(2-i)w=3+4i$$

$$iz+(3+i)w=-1+5i$$
Relevant Equations
Complex numbers
$$(1+i)z+(2-i)w=3+4i$$
$$iz+(3+i)w=-1+5i$$

ok, multiplying the first equation by##(1-i)## and the second equation by ##i##, we get,

$$2z+(1-3i)w=7+i$$
$$-z+(-1+3i)w=-5-i$$

adding the two equations, we get ##z=2##,
We know that, $$iz+(3+i)w=-1+5i$$
$$⇒2i+(3+i)w=-1+5i$$
$$w=\frac {-1+3i}{3+i}$$
$$w=\frac {(-1+3i)(3-i)}{(3+i)(3-i)}$$
$$w=i$$

There may be a different approach from this...
 
Last edited:
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I'm not sure there is anything easier.
 
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Well, I could debit something that is different/is not different but looks different and is probably not easier :smile::
Consider this a matrix equation ## Ax = y## with solution ##x = A^{-1}y##:$$
\begin{pmatrix} 1+i&2-i\\\phantom{1+}i&3+i\end {pmatrix}
\begin{pmatrix} z\\w\end {pmatrix}=\begin{pmatrix} \phantom{-}3+4i\\-1+5i\end {pmatrix}
\quad \Rightarrow \quad
\begin{pmatrix} z\\w\end {pmatrix} =
{1\over |A| }\begin{pmatrix} 3+i&-2+i\\\phantom{1}-i&\phantom{-}1+i\end {pmatrix}
\begin{pmatrix} \phantom{-}3+4i\\-1+5i\end {pmatrix}
$$With ## |A| =\det A = (1+i)(3+i)-i(2-i) = 1 + 2i\ ##we get $$
\begin{pmatrix} z\\w\end {pmatrix} = {1\over 1 + 2i }
\begin{pmatrix} (3+i)(3+4i)+(-2+i)(-1+5i) \\ \phantom 3\ - i\phantom )(3+4i)+( \phantom{-}1+i)(-1+5i) \end {pmatrix} = {1\over 1 + 2i }
\begin{pmatrix}\phantom{-} 2+4i\\-2+{\phantom 4}i \end {pmatrix} =
\begin{pmatrix}2\\i\end {pmatrix}$$
##\LaTeX## wise this is quite a compact solution :wink: .
Yours could also be slightly more compact if you realize that $$ followed by a carriage return and then $$ for a new displayed equation creates a lot of vertical white spacing. And the more so with two carriage returns (another empty line !).

But that's all beside the mathematical point, where you did just fine!

##\ ##
 
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BvU said:
##\LaTeX## wise this is quite a compact solution :wink: .
It still takes a while to typeset. *grumble*grumble*deletes*half*complete*post*grumble* :wink:
 
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Which is the carriage return syntax? ...you could share a simple math example "with" and "without" carriage return then i can adopt that immediately...
 
I think the point is that if you write Newton's second law says $$F=ma$$ and is the definition of "force". you get:
Newton's second law says $$F=ma$$and is the definition of "force".​
But I think you're inserting a blank line (or at least a new line) before and after the LaTeX delimiters, so you get
Newton's second law says​
$$F=ma$$​
and is the definition of "force".​
...which takes up a lot more room vertically.
 
Thanks, i will check on that Ibix...
 
This time I had to grmbl grmbl :biggrin:

And I also lost the part I did want to keep :mad:, namely:
I had fun discovering that googling (3+i)(3+4i)+(-2+i)(-1+5i)
(copied straight from the ##\TeX## source!) gives ##2+4i##
But horizontally aligning (-i)(3+4i)+(1+i)(-1+5i) was nightmarish indeed :wink:

@chwala: experiment !
 
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BvU said:
Yours could also be slightly more compact if you realize that $$ followed by a carriage return and then $$ for a new displayed equation creates a lot of vertical white spacing. And the more so with two carriage returns (another empty line !).
I rarely use $$ for this exact reason. I mostly use ##.
chwala said:
Which is the carriage return syntax? ...you could share a simple math example "with" and "without" carriage return then i can adopt that immediately...
Carriage return -- enter key
 
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  • #10
I prefer $$ ...the equations look like a piece of art...I will just need to work on the spacing.

## doesn't bring out the art and desired neatness...particularly on fraction type of equations...
 
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  • #11
Also Cramer's Rule, which I probably would have used. A slightly different take on the linear algebra of @BvU's post.
 
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DaveE said:
Also Cramer's Rule, which I probably would have used. A slightly different take on the linear algebra of @BvU's post.
Yap...Matrices in the making...
 
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chwala said:
I prefer $$ ...the equations look like a piece of art...I will just need to work on the spacing.

## doesn't bring out the art and desired neatness...particularly on fraction type of equations...
For fractions with ## , use \dfrac for a fraction rather than \frac .
 
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