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Is there some fast way to know the real part of a quotient?

$$\Re\left(\frac{z_1}{z_2}\right)$$

$z_i\in \mathbb{C}$

Git Gud
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Maths Student
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3 Answers3

5

Given $z_1=a+bi$ and $z_2=c+di$ then:$$\Re\left(\frac{z_1}{z_2}\right)=\frac{ac+bd}{c^2+d^2}$$

Eleven-Eleven
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user130512
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  • perfect...that should do it. – Eleven-Eleven Mar 10 '14 at 18:47
  • If we apply you result to: $\frac{e^{i\pi}-1}{e^{i\pi/(M+1)}-1}=\frac{-2}{cos \alpha + i sin \alpha -1}$ where $\alpha=\frac{\pi}{M+1}$ so we obtain that real part of this quotient is $\frac{-2cos \alpha}{cos^2 \alpha-2 cos\alpha +1 + sin^2 \alpha}=\frac{cos\alpha}{cos\alpha -1}?$ – Maths Student Mar 10 '14 at 18:57
  • @MathStudent: I think you obtain $\Re \left( \frac{e^{i \pi} - 1}{e^{i \alpha} - 1} \right) = \frac{-2(\cos \alpha - 1)}{(\cos \alpha - 1)^2 + \sin^2 \alpha} = 1$. – Paolo Jul 09 '15 at 08:27
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If you don't have $z_1$ and $z_2$ in a "nice" form eg some of the values of $e^{i\theta}$ for various $\theta$, you could use $1/2(z+\bar{z})=\Re{(z)}$.

So you'd get:

$\Re(\frac{z_1}{z_2})=\frac{z_1\bar{z_2}+\bar{z_1}z_2}{2z_2\bar{z_2}}$

for
$z_1=r_1e^{i\theta_1}$

$z_2=r_2e^{i\theta_2}$

$\Re(\frac{z_1}{z_2})=\frac{r_1}{r_2}cos(\theta_1-\theta_2)$

Just as an alternative, although I suspect you wanted something like user130512's post.

snulty
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At user130512's request in a comment, Euler's form. (I really suspect the calculation is going to get bad).

Let $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$.

Now $\frac{z_1}{z_2} = \frac{r_1}{r_2}e^{i(\theta_1 - \theta_2)}$

Assuming $z_1 = a + ib, z_2 = c+id$

$$\Re\left(\frac{z_1}{z_2}\right) = \frac{r_1}{r_2}\cos\left(\arctan\left(\frac{b}{a}\right) - \arctan\left(\frac{d}{c}\right)\right)$$

Disclaimer This is for pure amusement. No practical point in doing this.

Guy
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