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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

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Comment | Dear Prof Vijilant Thanks Prashant |

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