-1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Hint: Multiply the numerator and denominator by $c-di$. $\endgroup$
    – Rocket Man
    Dec 6, 2015 at 7:44

2 Answers 2

1
$\begingroup$

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 $-ad+bc=0$.

$\endgroup$
0
$\begingroup$

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

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

So:

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



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

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.