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When given a equation containing complex numbers such as

$$ \frac{a+ib}{c+id} = x + iy$$

and required to prove

$$ \frac{a^2 +b^2}{c^2+d^2} = x^2 + y^2$$

Is taking the mod of both sides a legal mathematical step? I ask so because my textbook first finds the conjugate of both sides and multiplies the conjugate with the initial equation. This is done everywhere a case like the above one occurs. Hence im a bit confused if the way im thinking of proceeding is mathematically legal.

noobAtMaths
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  • $$\left| \frac{z_1}{z_2} \right|=\frac{|z_1|}{|z_2|}$$ – Behrouz Maleki Jun 21 '16 at 19:20
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    It is fine, but we would need to know the theorem that the mod of a ratio is the ratio of the mods. Or else cross multiply, but we still have to know that the mod of a product is the product of the mods. – André Nicolas Jun 21 '16 at 19:21
  • Surely all this is obvious if you write $z=re^{i\theta}$ or simply as $r(\cos\theta+i\sin\theta)$ – almagest Jun 21 '16 at 19:25

2 Answers2

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You can perform any function you like to both sides of an equality and you will get an equality.

This is actually baked into the definition of a function $f\colon X \to Y$. If you give a function $f$ any $x$-value whatsoever, it is required to give you back a single value $f(x)$.

If you give a function $f$ two $x$-values that are in fact the same, $x_1 = x_2$, it is legally bound to give you back the same value for each; $f(x_1) = f(x_2)$.

Since the modulus function $$ x + iy \mapsto |x + iy| = \sqrt{x^2 + y^2} $$ is a function, it preserves equality. As does $\sqrt{\ \ \ }$, and $\sin(\ \ )$, and $\exp(\ \ )$, and the "divide by $2$" function, and ... I think you get the picture :)

PS, I think using the modulus is a very reasonable thing to do here, especially if you know $|z_1 / z_2| = |z_1|/|z_2|$.

pjs36
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Yes, taking the mod of both sides is mathematically valid, but we don't necessarily need to do that.

Convert to polar: $$\frac{\sqrt{a^2+b^2}\text{cis}(\theta_1)}{\sqrt{c^2+d^2}\text{cis}(\theta_2)}=\sqrt{x^2+y^2}\text{cis}(\theta_3)$$ Separate the moduli and arguments: $$\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}\sqrt{x^2+y^2}}=\frac{\text{cis}(\theta_3)\text{cis}(\theta_2)}{\text{cis}(\theta_1)}=\text{cis}(\theta_3+\theta_2-\theta_1)$$ The left-side is obviously a positive real number, but the only positive real number in the range of $\text{cis}$ is $1$, so we have: $$\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}\sqrt{x^2+y^2}}=1$$ Multiply both sides by $\sqrt{x^2+y^2}$: $$\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}=\sqrt{x^2+y^2}$$ Square both sides: $$\frac{a^2+b^2}{c^2+d^2}=x^2+y^2$$

Noble Mushtak
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