I Feel Nostalgic!

Algebra Level 3

Let $\{a,b\}$ and $\{m,n\}$ be consecutive terms (in that order) of 2 different geometric progressions with common ratio $r_1>0$ and $r_2>0$ respectively, such that $(a-b)m=(a+b)n.$ Let $z_1$ and $z_2$ be $2$ complex numbers such that $\begin{aligned} \arg(z_1z_2)&=\arctan(r_1)+\arctan(r_2)=\theta\quad\quad (0\leq \theta\leq \pi), \\ \arg\left(\frac{z_1}{z_2}\right)&=\frac{\pi}9. \end{aligned}$

Find $2\arg(z_1)$ .

Notation: The function $\arg(z)$ denotes the argument of a complex number.

Clarification: Take the value of $\pi=3.1415$ .

The answer is 1.134.

1 solution

Sravanth C.
Feb 16, 2017

We start with the fact that $b=ar_1$ and $n=mr_2$ , therefore $\begin{aligned} (a-b)m&=(a+b)n\\ am-amr_2&=amr_1+amr_1r_2\\ r_1+r_2&=1-r_1r_2\\ \frac{r_1+r_2}{1-r_1r_2}&=1 \end{aligned}$

Now observe the above expression, doesn't it look like $\tan(A+B)$ formula? Substituting $r_1=\tan A$ and $r_1=\tan B$ , we arrive at $\begin{aligned} \tan(A+B)&=1\\ A+B=\arctan r_1 +\arctan r_2 &=\arctan 1=\pi/4 \end{aligned}$

Now combining the result with the last statements we get $\begin{aligned} \arg(z_1z_2)&=\pi/4\\ \arg(z_1/z_2)&=\pi/9\\ \text{adding the above equations:}\quad\arg(z_1z_2)+\arg(z_1/z_2)&=\arg(z_1^2)\\ &=2\arg z_1\\ &=\pi/4+\pi/9\\ &=1.134 \end{aligned}$

Why did you state that $\theta \leq 0 \leq \pi$ ? Is that a typo?

Calvin Lin Staff - 4 years, 3 months ago

Yeah, sorry.

Sravanth C. - 4 years, 3 months ago

I Feel Nostalgic!

The answer is 1.134.

1 solution

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