If 13 e^{i tan ^{-1} frac{5}{12}} = a + i b,t | vedclass

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(A) Given,$13 e^{i 	an ^{-1} \frac{5}{12}} = a + i b$. Using Euler's formula $e^{i 	heta} = \cos 	heta + i \sin 	heta$,we have: $13 [\cos(	an ^{-

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$(12, 5)$

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(A) Given,$13 e^{i 	an ^{-1} \frac{5}{12}} = a + i b$. Using Euler's formula $e^{i 	heta} = \cos 	heta + i \sin 	heta$,we have: $13 [\cos(	an ^{-1} \frac{5}{12}) + i \sin(	an ^{-1} \frac{5}{12})] = a + i b$. Let $	heta = 	an ^{-1} \frac{5}{12}$,then $	an 	heta = \frac{5}{12}$. In a right-angled triangle with opposite side $5$ and adjacent side $12$,the hypotenuse is $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$. Thus,$\cos 	heta = \frac{12}{13}$ and $\sin 	heta = \frac{5}{13}$. Substituting these values: $13 [\frac{12}{13} + i \frac{5}{13}] = a + i b$. $12 + 5i = a + i b$. Comparing real and imaginary parts,we get $a = 12$ and $b = 5$. Therefore,the ordered pair $(a, b) = (12, 5)$.

If $13 e^{i \tan ^{-1} \frac{5}{12}} = a + i b$,then the ordered pair $(a, b) =$

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