Subfield: Algebra


Question: If $6 \tan ^{-1} x+4 \tan ^{-1}(3 x)=\pi$, compute $x^{2}$.

Solution:
$\quad$ Let $z=1+x i$ and $w=1+3 x i$, where $i=\sqrt{-1}$. Then $\tan ^{-1} x=\arg z$ and $\tan ^{-1}(3 x)=\arg w$, where $\arg z$ gives the measure of the angle in standard position whose terminal side passes through $z$. By DeMoivre's theorem, $6 \tan ^{-1} x=\arg \left(z^{6}\right)$ and $4 \tan ^{-1}(3 x)=\arg \left(w^{6}\right)$. Therefore the equation $6 \tan ^{-1} x+4 \tan ^{-1}(3 x)=\pi$ is equivalent to $z^{6} \cdot w^{4}=a$, where $a$ is a real number (and, in fact, $a<0$ ). To simplify somewhat, we can take the square root of both sides, and get $z^{3} \cdot w^{2}=0+b i$, where $b$ is a real number. Then $(1+x i)^{3}(1+3 x i)^{2}=$ $0+b i$. Expanding each binomial and collecting real and imaginary terms in each factor yields $\left(\left(1-3 x^{2}\right)+\left(3 x-x^{3}\right) i\right)\left(\left(1-9 x^{2}\right)+6 x i\right)=0+b i$. In order that the real part of the product be 0 , we have $\left(1-3 x^{2}\right)\left(1-9 x^{2}\right)-\left(3 x-x^{3}\right)(6 x)=0$. This equation simplifies to $1-30 x^{2}+33 x^{4}=0$, yielding $x^{2}=\frac{15 \pm 8 \sqrt{3}}{33}$. Notice that $\frac{15 \pm 8 \sqrt{3}}{33} \approx 1$, which would mean that $x \approx 1$, and $\operatorname{so} \tan ^{-1}(x) \approx \frac{\pi}{4}$, which is too large, since $6 \cdot \frac{\pi}{4}>\pi$. (It can be verified that this value for $x$ yields a value of $3 \pi$ for the left side of the equation.) Therefore we are left with $x^{2}=\frac{15-8 \sqrt{3}}{\mathbf{3 3}}$. To verify that this answer is reasonable, consider that $\sqrt{3} \approx 1.73$, so that $15-8 \sqrt{3} \approx 1.16$, and so $x^{2} \approx \frac{7}{200}=0.035$. Then $x$ itself is a little less than 0.2 , and so



$\tan ^{-1} x \approx \frac{\pi}{15}$. Similarly, $3 x$ is about 0.6 , so $\tan ^{-1}(3 x)$ is about $\frac{\pi}{6} \cdot 6 \cdot \frac{\pi}{15}+4 \cdot \frac{\pi}{6}$ is reasonably close to $\pi$.
Recall that $\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a \tan b}$, thus $\tan (2 a)=\frac{2 \tan a}{1-\tan ^{2} a}$ and

$$
\tan (3 a)=\tan (2 a+a)=\frac{\frac{2 \tan a}{1-\tan ^{2} a}+\tan a}{1-\frac{2 \tan a}{1-\tan ^{2} a} \cdot \tan a}=\frac{2 \tan a+\tan a-\tan ^{3} a}{1-\tan ^{2} a-2 \tan ^{2} a}=\frac{3 \tan a-\tan ^{3} a}{1-3 \tan ^{2} a}
$$

Back to the problem at hand, divide both sides by 2 to obtain $3 \tan ^{-1} x+2 \tan ^{-1}(3 x)=\frac{\pi}{2}$. Taking the tangent of the left side yields $\frac{\tan \left(3 \tan ^{-1} x\right)+\tan \left(2 \tan ^{-1}(3 x)\right)}{1-\tan \left(3 \tan ^{-1} x\right) \tan \left(2 \tan ^{-1}(3 x)\right)}$. We know that the denominator must be 0 since $\tan \frac{\pi}{2}$ is undefined, thus $1=\tan \left(3 \tan ^{-1} x\right) \tan \left(2 \tan ^{-1}(3 x)\right)=$ $\frac{3 x-x^{3}}{1-3 x^{2}} \cdot \frac{2 \cdot 3 x}{1-(3 x)^{2}}$ and hence $\left(1-3 x^{2}\right)\left(1-9 x^{2}\right)=\left(3 x-x^{3}\right)(6 x)$. Simplifying yields $33 x^{4}-$ $30 x^{2}+1=0$, and applying the quadratic formula gives $x^{2}=\frac{15 \pm 8 \sqrt{3}}{33}$. The " + " solution is extraneous: as noted in the previous solution, $x=\frac{15+8 \sqrt{3}}{33}$ yields a value of $3 \pi$ for the left side of the equation), so we are left with $x^{2}=\frac{\mathbf{1 5}-\mathbf{8} \sqrt{\mathbf{3}}}{\mathbf{3 3}}$.

Final answer: $\frac{15-8 \sqrt{3}}{33}$