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# Number Theory/Compute Logarithm and Antilogarithm of complex numbers.

Question
Dear Prof Vijilant

Can we compute the Logarithm and Antilogarithm of complex numbers?

Examples

Log(100+2i)
Antilog(3-4i)

Thanks
Prashant

Yes Prashant
I'll assume you mean log to base 10
log(100 + 2i) = ln(100 + 2i)/ln(10)
arg(100 + 2i) = invtan(1/50).  mod(100 + 2i) = (10004)^(1/2)
So log(100 + 2i) = (1/2ln(10004) + i*invtan(1/50))/ln(10)

antilog(3 - 4i) = 10^(3 - 4i) = 1000*10^(-4i).  Put 10^(-4i) = z, then ln(z) = -4i*ln(10)
And z = e^(-4ln(10)i) = cos(-4ln(10)) + isin(-4ln(10))
So antilog(3 - 4i) = 1000*(cos(4ln(10)) - isin(4ln(10)))

Best wishes
vijilant
Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Dear Prof Vijilant Thanks Prashant

Number Theory

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#### Vijilant

##### Expertise

Most questions on number theory, divisibility, primes, Euclidean algorithm, Fermat`s theorem, Wilson`s theorem, factorisation, euclidean algorithm, diophantine equations, Chinese remainder theorem, group theory, congruences, continued fractions.

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Teacher of math for 53 years

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I taught John Birt, former Director of the BBC in 1961. His homework book was the most perfect I have ever marked. And also the most neat. I could tell he was destined for great things. One of my classmates was the poet Roger McGough, and I have a mention in his autobiography.