Consider a ratio having complex numerator and denominator $(a+bi)$/$(c+di)$. when will this ratio become real ? the obvious answer is when imaginary part is 0. but when will it become real?

  • $\begingroup$ Hint: Multiply the numerator and denominator by $c-di$. $\endgroup$
    – Rocket Man
    Dec 6, 2015 at 7:44

2 Answers 2


As you say, when the imaginary part is $0$: $\frac{a+bi}{c+di}=\frac{(a+bi)(c-di)}{c^2+d^2}=\frac{ac+bd}{c^2+d^2}+\frac{-ad+bc}{c^2+d^2}i$. So, this will be real when the imaginary part is $0$, namely, when $-ad+bc=0$.


Assume $a,b,c,d\in\mathbb{R}$:



$$\Re\left(z\right)=\Re\left(\frac{a+bi}{c+di}\right)=\Re\left(\frac{ac+bd}{c^2+d^2}-\frac{ad-bc}{c^2+d^2}i\right)=\frac{ac+bd}{c^2+d^2}$$ $$\Im\left(z\right)=\Im\left(\frac{a+bi}{c+di}\right)=\Im\left(\frac{ac+bd}{c^2+d^2}-\frac{ad-bc}{c^2+d^2}i\right)=-\frac{ad-bc}{c^2+d^2}$$

If you want that your $z$ is totally real then the imaginary part of $z$ has to be zero:

$$\Im\left(\frac{a+bi}{c+di}\right)=0\Longleftrightarrow$$ $$-\frac{ad-bc}{c^2+d^2}=0\Longleftrightarrow$$ $$\frac{ad-bc}{c^2+d^2}=0\Longleftrightarrow$$ $$\frac{ad-bc}{c^2+d^2}=\frac{0}{1}\Longleftrightarrow$$ $$ad-bc=0$$

  • $\begingroup$ In the first line, the numerator $ac+bd+i(ad-bc)$ should be $ac+bd-i(ad-bc)$. $\endgroup$
    – Wojowu
    Dec 6, 2015 at 10:26
  • $\begingroup$ @Wojowu Thanks for your feedback, I had made a mistake! $\endgroup$ Dec 6, 2015 at 10:41

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