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Number Theory/Compute Logarithm and Antilogarithm of complex numbers.

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Question
Dear Prof Vijilant

Can we compute the Logarithm and Antilogarithm of complex numbers?

Examples

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

Thanks
Prashant

Answer
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  

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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.

Experience

Teacher of math for 53 years

Organizations
AQA Doncaster Bridge Club Danum Strings Orchestra Doncaster Conservative Club Danum Strings Orchestra Simply Voices Choir Doncaster TNS mystery shopping St Paul's Music Group Cantley

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Journal of mathematics and its applications M500 magazine

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BSc (Hons) Liverpool (Science). BA (Hons) OU (Mathematics)

Awards and Honors
State Scholarship 1955 Highest Score in Yorkshire on OU course MST209 50 prize First class honours in OU BA Mathematics

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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.

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