Complex number is complex

Algebra Level pending

ln ⁡ ( 3 + 4 i ) = A B ln ⁡ C + i ( 2 n π + arctan ⁡ ( D E ) ) \ln(3+4i) = \dfrac AB \ln C + i \left(2n \pi + \arctan\left( \dfrac DE \right) \right)

If A , B , C , D A,B,C,D and E E are positive integers satisfying the equation above, with gcd ⁡ ( A , B ) = gcd ⁡ ( D , E ) = 1 \gcd(A,B) = \gcd(D,E) = 1 , find A + B + C + E A+B+C+E .


The answer is 31.

Rishabh Deep Singh by Rishabh Deep Singh , 23, India

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1 solution

Tom Engelsman
Apr 26, 2020

We can write the above complex number in its Eulerian form: 3 + 4 i = 25 e i arctan ⁡ ( 4 / 3 ) 3+4i = \sqrt{25}e^{i \arctan(4/3)} . Taking the logarithm of this expression produces:

l n ( 25 e i arctan ⁡ ( 4 / 3 ) ) = 1 2 ln ⁡ ( 25 ) + i ( arctan ⁡ ( 4 3 ) ) ln(\sqrt{25}e^{i \arctan(4/3)}) = \frac{1}{2} \ln(25) + i(\arctan(\frac{4}{3})) .

Thus, A + B + C + E = 1 + 2 + 25 + 3 = 31 . A +B + C + E = 1+2+25+3 = \boxed{31}.

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