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I have recently started a problem related to complex numbers. The topic is new to me, but I understand how do they work though. However there is one concept which is a little bit "intricate" to me, and this is the square of the magnitude of a complex number. I know how the square of the magnitude of a single complex number looks like, but what if i have an expression? Here's what I mean:

$$|a+b+2ab|^2$$

where $a$ and $b$ are complex numbers whose magnitude are 1.

Let's say $a=x_1+y_1i$ and $b=x_2+y_2i$, so the expression should be $$|x_1+y_1i+2(x_1+y_1i)(x_2+y_2i)+x_2+y_2i|^2$$

which is

$$|x_1+y_1i+x_2+y_2i+2x_1x_2+2x_1y_2i+2x_2y_1i-2y_1y_2|^2$$

right?

So what I don't know is how to use the magnitude here. Should the real parts and imaginary parts be squared as a whole respectively, or each term should be squared?

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    $\begingroup$ You should group the imaginary and real terms together then treat it as a single number. $\endgroup$ Commented Jan 12 at 17:15
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    $\begingroup$ As CyclotomicField said, the last term you obtain is in itself a third complex number. Obtain its magnitude, then square the magnitude. $\endgroup$
    – zxayn
    Commented Jan 12 at 17:17
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    $\begingroup$ I suspect $|a + b +2ab|^2$ was a biproduct of another question. What are you really trying to do that this came up? $\endgroup$
    – fleablood
    Commented Jan 12 at 17:18
  • $\begingroup$ @fleablood it is a part of an inequality, it has three terms, similar to this, but they are symmetrical. That's why it is presented such way. $\endgroup$
    – fikooo
    Commented Jan 12 at 17:23
  • $\begingroup$ You should get used to thinking in terms if $a= x_r +x_i i$ and $b=y_r + y_i i$ that $a + b = (x_r + y_r) + (x_i + y_i)i$ and $ab = (x_ry_r- x_iy_i) + (x_iy_r + x_ry_i)i$ and so $|a + b + 2ab|= |(x_r + y_r + 2 (x_ry_r- x_iy_i)) + (x_i + y_i + 2(x_iy_r + x_ry_i))i| = \sqrt{(x_r + y_r + 2 (x_ry_r- x_iy_i))^2 + (x_i + y_i + 2(x_iy_r + x_ry_i))^2}$ while bearing in mind that $x_r^2 + x_i^2 =1 = y_r^2 + y_i^2$.... Not that I recommend you actually follow through on this.... $\endgroup$
    – fleablood
    Commented Jan 12 at 17:25

2 Answers 2

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HINT

Since $|a| = |b| = 1$, we may express them as $a = e^{i\theta}$ and $b = e^{i\varphi}$. Consequently, one has that: \begin{align*} |a + b + 2ab|^{2} & = (a + b + 2ab)(\overline{a} + \overline{b} + 2\overline{ab})\\\\ & = |a|^{2} + a\overline{b} + 2|a|^{2}\overline{b} + \overline{a}b + |b|^{2} + 2\overline{a}|b|^{2} + 2|a|^{2}b + 2a|b|^{2} + 4|a|^{2}|b|^{2}\\\\ & = 1 + a\overline{b} + 2\overline{b} + \overline{a}b + 1 + 2\overline{a} + 2b + 2a + 4\\\\ & = 6 + (a\overline{b} + \overline{a}b) + 2(a + \overline{a}) + 2(b + \overline{b}) \end{align*}

Can you take it from here?

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  • $\begingroup$ The furthest I got is $$6+2x_1x_2+2y_1y_2+4x_2+4x_1$$ using the notation I've used. Now I'm stuck again. $\endgroup$
    – fikooo
    Commented Jan 12 at 18:09
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From $$|x_1+y_1i+x_2+y_2i+2x_1x_2+2x_1y_2+2x_2y_1-2y_1y_2|^2$$ regroup as $$|(x_1+x_2+2x_1x_2+2x_1y_2+2x_2y_1-2y_1y_2) + (y_1+y_2)i|^2$$ which in turn will be $$(x_1+x_2+2x_1x_2+2x_1y_2+2x_2y_1-2y_1y_2)^2 + (y_1+y_2)^2$$ and notice that you can always regroup the terms to have a real an imaginary part. Another way you might try to calculate such an expression is to note that $|z|^2 = z\overline{z}$ where $z=a+bi$ and $\overline{z}=a-bi$, the conjugate of $z$. This can sometimes help avoid breaking each term up into real and imaginary parts.

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