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Show that
$$\tan \left(z_{1}+z_{2}\right)=\frac{\tan z_{1}+\tan z_{2}}{1-\left(\tan z_{1}\right)\left(\tan z_{2}\right)}$$
for all complex numbers $z_{1}$ and $z_{2}$ satisfying $z_{1}, z_{2}, z_{1}+z_{2} \neq n \pi+\pi / 2$ for any integer
$n$
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## 1 Answer

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Proof. Since
\begin{aligned} & \tan z_{1}+\tan z_{2}=\frac{i\left(e^{-i z_{1}}-e^{i z_{1}}\right)}{e^{i z_{1}}+e^{-i z_{1}}}+\frac{i\left(e^{-i z_{2}}-e^{i z_{2}}\right)}{e^{i z_{2}}+e^{-i z_{2}}} \\ =& i \frac{\left(e^{-i z_{1}}-e^{i z_{1}}\right)\left(e^{i z_{2}}+e^{-i z_{2}}\right)+\left(e^{-i z_{2}}-e^{i z_{2}}\right)\left(e^{i z_{1}}+e^{-i z_{1}}\right)}{\left(e^{i z_{1}}+e^{-i z_{1}}\right)\left(e^{i z_{2}}+e^{-i z_{2}}\right)} \\ =&-2 i \frac{e^{i\left(z_{1}+z_{2}\right)}-e^{-i\left(z_{1}+z_{2}\right)}}{\left(e^{i z_{1}}+e^{-i z_{1}}\right)\left(e^{i z_{2}}+e^{-i z_{2}}\right)} \end{aligned}
and
\begin{aligned} & 1-\left(\tan z_{1}\right)\left(\tan z_{2}\right)=1-\left(\frac{i\left(e^{-i z_{1}}-e^{i z_{1}}\right)}{e^{i z_{1}}+e^{-i z_{1}}}\right)\left(\frac{i\left(e^{-i z_{2}}-e^{i z_{2}}\right)}{e^{i z_{2}}+e^{-i z_{2}}}\right) \\ =& \frac{\left(e^{-i z_{1}}+e^{i z_{1}}\right)\left(e^{-i z_{2}}+e^{i z_{2}}\right)+\left(e^{-i z_{1}}-e^{-i z_{1}}\right)\left(e^{-i z_{2}}-e^{i z_{2}}\right)}{\left(e^{-i z_{1}}+e^{i z_{1}}\right)\left(e^{-i z_{2}}+e^{i z_{2}}\right)} \\ =& 2 \frac{e^{i\left(z_{1}+z_{2}\right)}+e^{-i\left(z_{1}+z_{2}\right)}}{\left(e^{i z_{1}}+e^{-i z_{1}}\right)\left(e^{i z_{2}}+e^{-i z_{2}}\right)} \end{aligned}

we have
$$\frac{\tan z_{1}+\tan z_{2}}{1-\left(\tan z_{1}\right)\left(\tan z_{2}\right)}=-i \frac{e^{i\left(z_{1}+z_{2}\right)}-e^{-i\left(z_{1}+z_{2}\right)}}{e^{i\left(z_{1}+z_{2}\right)}+e^{-i\left(z_{1}+z_{2}\right)}}=\tan \left(z_{1}+z_{2}\right)$$

Alternatively, we can argue as follows if we assume that the identity holds for $z_{1}$ and $z_{2}$ real. Let
$$F\left(z_{1}, z_{2}\right)=\tan \left(z_{1}+z_{2}\right)-\frac{\tan z_{1}+\tan z_{2}}{1-\left(\tan z_{1}\right)\left(\tan z_{2}\right)}$$
We assume that $F\left(z_{1}, z_{2}\right)=0$ for all $z_{1}, z_{2} \in \mathbb{R}$ with $z_{1}, z_{2}, z_{1}+z_{2} \neq n \pi+\pi / 2$.
Fixing $z_{1} \in \mathbb{R}$, we let $f(z)=F\left(z_{1}, z\right)$. Then $f(z)$ is analytic in its domain
$$\mathbb{C} \backslash\left(\{n \pi+\pi / 2\} \cup\left\{n \pi+\pi / 2-z_{1}\right\}\right)$$
And we know that $f(z)=0$ for $z$ real. Therefore, by the uniqueness of analytic functions, $f(z) \equiv 0$ in its domain. So $F\left(z_{1}, z_{2}\right)=0$ for all $z_{1} \in \mathbb{R}$ and $z_{2} \in \mathbb{C}$ in its domain. Fixing $z_{2} \in \mathbb{C}$, we let $g(z)=F\left(z, z_{2}\right) .$ Then $g(z)$ is analytic in its domain
$$\mathbb{C} \backslash\left(\{n \pi+\pi / 2\} \cup\left\{n \pi+\pi / 2-z_{2}\right\}\right)$$
And we have proved that $g(z)=0$ for $z$ real. Therefore, by the uniqueness of analytic functions, $g(z) \equiv 0$ in its domain. Hence $F\left(z_{1}, z_{2}\right)=0$ for all $z_{1} \in \mathbb{C}$ and $z_{2} \in \mathbb{C}_{\text {in }}$ its domain.
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