I'm starting to read a book on complex analysis, and I'm having some troubles envolving simple equations with complex numbers. How can I solve equations envolving these numbers- what methods and strategies do you recommend? When it envolves aspects like $\bar z$, $|z|$ and $Arg(z)$, what should I do? Please take this equation as an example:$$|z|-z=1+2i$$

  • $\begingroup$ Try replacing $z$ by $x +iy$. Also, remember that for complex numbers to be equal, their real and imaginary parts must be equal. $\endgroup$ – manthanomen Apr 9 '17 at 4:16
  • $\begingroup$ I tried it the first time, but then got stuck with $\sqrt {x^2+y^2}-x+iy=1+2i$. $\endgroup$ – TPace Apr 9 '17 at 4:23
  • $\begingroup$ Now break it into two real equations. $\endgroup$ – Jean-François Gagnon Apr 9 '17 at 4:24
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    $\begingroup$ The $+$ on the left side of your equation should be a $-$. Now what you want to do is set up two equations (one for the real part and one for the imaginary part). $\endgroup$ – manthanomen Apr 9 '17 at 4:26
  • $\begingroup$ What do you mean by setting up a real and imaginary part equations? What should they be like? $\endgroup$ – TPace Apr 9 '17 at 4:30

For your specific question, remember that $z$ can be written as $z=x+iy$ and $|z|=\sqrt{x^2+y^2}$ is a real number. Also, $a+bi=x+yi$ if and only if $a=x$ and $b=y$ at the same time.

There are some handy tricks when dealing with complex number equations. Many of them will involve replacing $z$ with one of these:

  • $z=x+iy$
  • $z=|z|(\cos\theta+i\sin\theta)$
  • $z=|z|e^{i\theta}$

where $\theta=Arg(z)$. Then you will be able to apply some formulas and identities and to get your result.

Sometimes it may be useful to think about it as a point in the complex plane. Then the trigonometric representation will make more sense.

Other times, when dealing with more analytical expressions (such as complex polynomials), it may be better to think in terms of the exponential.

Sometimes it will suffice to compare the real and imaginary terms (as in your exercise).

In the end, it all comes down to practice and familiarization. Do many exercises, work out some proofs in your books and try to work with them in other subjects, like Linear Algebra. Once you do, you'll have one of the most powerful tools of mathematics and physics at your disposal.

  • $\begingroup$ Thank you for your answer, I found it helpful and will use it in future exercises. I've just one question: is $\theta$ equal to $Arg(z)$ in your case? $\endgroup$ – TPace Apr 9 '17 at 4:42
  • $\begingroup$ Yes, it is. I prefer writing $\theta$ to shorten it :) $\endgroup$ – AspiringMathematician Apr 9 '17 at 4:43
  • $\begingroup$ Isn't $\theta =arctan(\frac{y}{x})$ just applicable when $x>0$? I read that, depending on the values of $x$ and $y$, sometimes you add or subtract $\pi$. Please answer a devoted youngling :v. $\endgroup$ – TPace Apr 9 '17 at 4:47
  • $\begingroup$ Actually, you're right. It's usually defined such that $\theta \in (-\pi,pi]$ or $\theta \in [0,2\pi)$, depending on convenience. I will edit the answer to define $\theta$ as $Arg(z)$ to avoid confusion. $\endgroup$ – AspiringMathematician Apr 9 '17 at 4:57
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    $\begingroup$ Yes, one of the things that interests me the most in complex numbers is its geometrical representation. Thank you by your answers, they were- and shall be- much helpful! $\endgroup$ – TPace Apr 9 '17 at 5:04

Let $z = x + iy$. Rewrite your equation as $|z| = 1 + 2i + z$. Since $|z|$ is a real number, we conclude that the imaginary part of $z$ must be $-2i$. So the equation we're really trying to solve is $$\sqrt{x^2 + 4} = 1 + x$$ This is equivalent to $$x^2 + 4 = 1 + 2x + x^2$$ i.e., $x = \frac{3}{2}$.

So the only solution is $z = \frac{3}{2} - 2i$.

  • $\begingroup$ I completely understand your logic, except by the part of why Im(z)=-2i. Sorry if my mathematical induction is failing, but I can't see how got there. $\endgroup$ – TPace Apr 9 '17 at 4:36
  • $\begingroup$ Since the modulus $|z|$ has to be a real number, we know the imaginary part $iy$ of $z$ must cancel out the $2i$ on the right hand side, so we must have $y = -2$. $\endgroup$ – manthanomen Apr 9 '17 at 4:56
  • $\begingroup$ Oooooh, now I understood you. I'll try to apply the same logic to other questions. Thank you a lot! $\endgroup$ – TPace Apr 9 '17 at 5:03

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