If $13 e^{i \tan ^{-1 \frac{5}{12}}}=a+i b$, then the ordered pair $(\mathrm{a}, \mathrm{b})=$
If $13 e^{i \tan ^{-1 \frac{5}{12}}}=a+i b$, then the ordered pair $(\mathrm{a}, \mathrm{b})=$
- $(12,5)$
- $(5,12)$
- $(24,10)$
- $(10,24)$
Solution
We have,
$
\begin{gathered}
13 e^{i \tan ^{-1} \frac{5}{12}}=a+i b \\
\Rightarrow 13 \cdot\left[\cos \left(\tan ^{-1} \frac{5}{12}\right)+i \sin \left(\tan ^{-1} \frac{5}{12}\right)\right]=a+i b
\end{gathered}
$
$
\begin{array}{rlrl}
& \Rightarrow 13\left[\cos \left(\cos ^{-1} \frac{12}{13}\right)+i \sin \left(\sin ^{-1} \frac{5}{13}\right)\right]=a+i b \\
& \Rightarrow & 13\left[\frac{12}{13}+i \frac{5}{13}\right] & =a+i b \\
& \Rightarrow & 12+5 i & =a+i b \\
& \therefore & a & =12, b=5 \\
& \therefore & (a, b) & =(12,5)
\end{array}
$
Asked in: AP EAMCET 2018 (22 Apr Shift 1)
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