For the following equation:

$\frac{Aj}{100\sqrt{2}} -\frac{A}{100\sqrt{2}} +\frac{x}{200}-\frac{xj}{200} =0$ where $A$ is a constant and $j$ is the imaginary unit.

I thought the solution would be found by isolating the real and imaginary parts and setting them both equal to zero. I would then get the answer $\frac{2A}{\sqrt{2}}+\frac{2A}{\sqrt{2}}j$, when the answer is $\frac{2A}{\sqrt{2}}$ (This is obvious by just plugging this solution in). Why does setting the real and imaginary parts to the real and imaginary parts of the right side, which are both zero, produce the wrong answer?

Thank you.


You have:$$\frac{Aj}{100\sqrt{2}} -\frac{A}{100\sqrt{2}} +\frac{x}{200}-\frac{xj}{200} =0$$$$\therefore\left(\frac{A}{100\sqrt{2}}-\frac{x}{200}\right)j+\left(\frac{x}{200}-\frac{A}{100\sqrt{2}}\right) =0$$Both real and imaginary components must equal zero which both lead to the answer:$$\frac{x}{200}=\frac{A}{100\sqrt{2}}$$Which simplifies to your desired answer.

In general, if:$$a+bj=0$$Then this implies that $a=0$ and $b=0$

Even more generally, if:$$\color{red}{a}+\color{blue}{b}j=\color{red}{c}+\color{blue}{d}j$$Then this implies that $\color{red}{a}=\color{red}{c}$ and $\color{blue}{b}=\color{blue}{d}$

  • $\begingroup$ But how would one get a complex $x$ from setting two "real-valued" equations to zero? I thought the complex part set to zero gives the complex part of the solution and the real equation gives the real-part of the solution. Thank you. $\endgroup$ – dylan7 May 5 '15 at 20:01
  • $\begingroup$ Note that in your equation the magnitude of both the real and imaginary parts are identical. i.e. your equation is equivalent to:$$a+aj=0$$Which has the solution $a=0$ where:$$a=\frac{A}{100\sqrt{2}}-\frac{x}{200}$$ $\endgroup$ – Mufasa May 5 '15 at 20:05
  • $\begingroup$ Ah ok thank you, but if they weren't equal $a+bj=0$, where $a$ is not the same equation as $b$, i.e. the real and imaginary parts are not identical, the solution would be $x_a+j*x_b$ where $x_a$ makes $a=0$ and $x_b$ makes $b=0$? $\endgroup$ – dylan7 May 5 '15 at 20:13
  • $\begingroup$ Not quite, if you had $a+bj=0$ where $a\ne b$ then the solution would be $a=0$ and $b=0$ as pointed out above in my answer. To take a specific example, lets say you had:$$(x-3)+(y-5)j=0$$The solutions to this would be $x=3$ and $y=5$. You cannot then go on to combine these and say that the solution is $3+5j$. All you are supposed to do is find the values of any variables within the real and imaginary parts of your equation that would make each of these components equal to zero. $\endgroup$ – Mufasa May 5 '15 at 20:16
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    $\begingroup$ If you had the equation:$$\left(\frac{A}{200}-\frac{x}{200}\right)j+\left(\frac{x}{200}-\frac{A\sqrt{3}}{200}\right) =0$$Then this leads to $x=A$ and $x=A\sqrt{3}$. Since $x$ can only take on one value, then this would only be valid if $A=0$ leading to $x=0$ for both solutions. If you had an equation that ended up with a solution like $x=1$ and $x=2$ then this would imply that there are no valid solutions to the equation. $\endgroup$ – Mufasa May 5 '15 at 20:35

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