# What is the quickest way of solving the system of linear complex equations?

I have the following system: $$x\left(-50-i\frac{2000}{3}\right)+y\left(-50-i\frac{1000}{3}\right)=-1$$ $$x\left(-50+i\frac{2000}{3}\right)+y\left(-50-i\frac{2000}{3}\right)=-1$$ $x,y\in\mathbb C.$

The method described here

is the only method that I know for solving the system of linear complex equations, but it is too long.

Question: Could someone show the quickest method (or quicker than this)?

Solution for this example is $$x=\frac{1}{16250}-i\frac{4}{8125},y=\frac{2}{8125}-i\frac{16}{8125}.$$

• Cramer's rule – Sri-Amirthan Theivendran Aug 6 '16 at 21:54
• Well, for starters, multiply each equation by $\dfrac{3}{50}$. – John Wayland Bales Aug 6 '16 at 22:05
• @Foobaz John What if the system is not squared? – user300046 Aug 6 '16 at 22:07

The values of coeficients are simetric. If you substract the second from the first equation you will get $4x - y = 0$.

Using that, just put $x= a +bi$ and $y= 4a + 4bi$ in any of the two original equations. The rest is easy, it is a system of real linear equations, and you can find $a, b$.

Consider the more general case:

$$ax+by=c$$

$$a'x+b'y=c'$$

You want to solve this, for some complex numbers $x,y$, where $a,b,c,a',b',c'$ are also complex numbers.

The easiest way is to use these variables to solve using substitutions probably. (Plug in the values of $a,b,c,\dots$ after you solve, then simplify if necessary)

In the more general case of

$$a_0x_0+a_1x_1+a_2x_2+\dots a_nx_n=y_0$$

$$b_0x_0+b_1x_1+b_2x_2+\dots b_nx_n=y_1$$

$$\vdots$$

$$k_0x_0+k_1x_1+k_2x_2+\dots k_nx_n=y_n$$

where you are solving for $x_0,x_1,x_2,\dots$, use Cramer's rule.