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I was wondering how to solve a linear equation of the below type.

In the below equation, z=x+yi, a complex number.

$$(2+2i)z^*+(2-2i)z=10+(2-2i)(2+2i)$$

So, in the above equation, there are two unknown variables and only one equation. I have learned about normal con-Sylvester matrix equation, but they don't seem to fit this problem.

I understand that it could be like: $$2*Re((2+2i)z^*)=10+(2-2i)(2+2i)$$

But the problem is how to find the value of x and y individually.

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2 Answers 2

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Let $z=x+yi.$ The equation $(2+2i)z^*+(2-2i)z=10+(2-2i)(2+2i)=10+4+4=18$

then becomes $(2+2i)(x-yi)+(2-2i)(x+yi)$

$=(2x+2y+(2x-2y)i)+(2x+2y+(2y-2x)i)=4x+4y=18,$

so it is solved when $x+y=\dfrac{18}4=\dfrac9 2$.

There are infinitely many solutions.

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Hint:

I suggest that you write $z$ as $x + i y$ and then expand all products.

Afterwards, split the equations in two parts: the real part and the imaginary part. This will leave you with a system of two equations with two unknowns.


Edit: Just for the sake of completeness, and after NIT_GUP solved the problem, here is the solution:

$$ (2+2i)z^∗ + (2−2i)z = 10 + (2−2i)(2+2i) \\ (2+2i)(x - i y) + (2−2i)(x + i y) = 10 + (2^2 + 2^2) \\ (2x \color{red}{+ 2ix} \color{blue}{- 2iy} + 2y) + (2x \color{red}{- 2ix} \color{blue}{+ 2iy} + 2y) = 18 \\ 4x + 4y = 18 \\ x + y = \frac{9}{2} $$

So, in this case we have a degenerate solution because the complex terms cancelled out in the equations above. In this case, the solution is a straight line in the complex plane with the equation: $$ y = - x + \frac{9}{2} \ . $$

plot of the equation y = - x + 9/2

Only after writing this I've noticed that J. W. Tanner had already written a solution.

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  • $\begingroup$ I tried, it worked. $\endgroup$
    – NIT_GUP
    Jun 27, 2019 at 23:35
  • $\begingroup$ I'm glad I could help. I added the solution in case somebody else finds a similar problem. If you think it's appropriate, you may accept one of the answers (there is a green button). $\endgroup$ Jun 28, 2019 at 12:05

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