2
$\begingroup$

The number $[\text{cis } 75^\circ + \text{cis } 83^\circ + \text{cis } 91^\circ + \dots + \text{cis } 147^\circ]$ is expressed in the form $r \, \text{cis } \theta$, where $0 \le \theta < 360^\circ$. Find $\theta$ in degrees.

How would I start this problem? I am sort of stuck on it.

Thanks

$\endgroup$

3 Answers 3

1
$\begingroup$

Firstly, you can begin by using Euler's formula to rewrite the series.

$$ \left[ e^{i 75^o} + e^{i 83^o} + \cdots + e^{i 147^o} \right] = e^{i 75^o} \left[ 1 +e^{i 8^o} + e^{i 16^o} + \cdots + e^{i 72^o} \right] $$

Or

$$ e^{i 75^o} \left[ 1 +e^{i 8^o} + (e^{i 8^o})^2 + \cdots + (e^{i 8^o})^9 \right] = \Sigma_{n=0}^9 e^{i 75^o} (e^{i 8^o})^n $$

Which is a geometric series with a common ratio of $e^{i 8^o}$ and a first term of $a= e^{i 75^o}$, which can be computed as

$$ \Sigma_{n=0}^{n-1} ar^n = a \frac{1-r^n}{1-r}$$

So we can plug in our values for $a$, $r$, and $n$.

$$ S = e^{i 75^o} \frac{1-(e^{i 8^o})^{10}}{1- e^{i 8^o}} = e^{i 75^o} \frac{1-e^{i 80^o}}{1- e^{i 8^o}}$$

Now is when we have to get a little more clever to return to $r \textrm{cis}{\theta}$ form, so we'll rewrite the series as

$$ \frac{\textrm{cis} 75^o - \textrm{cis} 155^o}{ 1 - \cos{8^o} -i \sin{8^o}}$$

Next, we have to multiply the numerator and denominator by the conjugate of the number in the denominator.

$$ \frac{\textrm{cis} 75^o - \textrm{cis} 155^o}{ 1 - \cos{8^o} -i \sin{8^o}} \frac{1-\cos{8^o} + i \sin{8^o}}{1-\cos{8^o} + i \sin{8^o}} $$

This will give us a completely real denominator. And from there, we just have a lot of algebra to grind out to get the remaining expression down to a form of $r \textrm{cis}{\theta}$. I hope you find the algebra to be doable.

$\endgroup$
0
$\begingroup$

The problem is basically a vector addition problem. $cis\theta = cos\theta + i sin\theta$. So, $cis 75 + cis 83 + ... = cos 75 + cos 83 + ... + i(sin 75 + sin 83 + ...)$

Actually doing the math would be difficult, but note that you're adding a bunch of vectors of equal length, each 8 degrees apart. They form an arc and what you want is the average direction of the arc since that is also the net direction of the vectors.

Hope that gets you started. I can comment further if necessary.

$\endgroup$
0
$\begingroup$

this is a geometric series $$cis(75^\circ) +cis(75^\circ + 8^\circ)+\cdots + cis(75^\circ+9\times 8^\circ)=cis(75^\circ)\frac{1-cis(80^\circ)}{1-cis(8^\circ)} $$ i hope you can take it from here.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .