How would one solve a complex equation system solely using a cartesian representation of complex numbers by hand? For instance, take the following linear equation system:
$(1+i)z_1 - z_2 = i$
$(1-i)z_1 + (1+i)z_2 = 1$
This system contains both complex variables and complex coefficients. Is it even possible to solve an equation system like this? How would one go about it? Does it have multiple solutions since it includes complex numbers?
$\endgroup$ 43 Answers
$\begingroup$Matrices with complex entries can be solved just as matrices with real entries.
$$ \left[ \begin{array}{cc|c} 1+i&-1&i\\ 1-i&1+i&1 \end{array} \right] $$
Reduce the matrix to RREF, using the exact same methods as real matrices. The only difference will be that you may have to do a bit of complex arithmetic.
In this case, you should end up with something like:
$$ \left[ \begin{array}{cc|c} 1&0&i/(i+1)\\ 0&1&0 \end{array} \right] $$
$\endgroup$ $\begingroup$$(1+i)z_1-z_2 = i\\ (1-i)z_1 + (1+i) = 1$
multiply the first equation by $(1+i)$, and add the two equations together.
$(1+i)^2z_1-(1+i)z_2 = i(1+i)\\ (1-i)z_1 + (1+i)z_2 = 1$
multiply out the complex numbers
$2i z_1-(1+i)z_2 = -1+i\\ (1-i)z_1 + (1+i)z_2 = 1$
adding the two equations
$(1+i)z_1 = i\\ z_1 = \frac {i}{1+i}$
multiply top and bottom by the complex conjugate
$z_1 = \frac {1 + i}2$
substitute into the original equation:
$(1+i)\frac {1 + i}2-z_2 = i\\ i-z_2 = i\\ z_2 = 0$
$\endgroup$ $\begingroup$Hint: treat $z_1$ and $z_2$ like normal real variables and solve for them. Here if you multiply the first equations by (1+i) you will get:
$(1+i)^2z_1 - (1+i)z_2 = i(1+i)$
$(1-i)z_1 + (1+i)z_2 = 1$
Add both equations:
$$\left[(1+i)^2+(1-i) \right]z_1=1+i(1+i) \implies z_1 = \frac{1+i(1+i)}{\left[(1+i)^2+(1-i) \right]}.$$
Plug this into one of the equations and solve for $z_2$.
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