Fnding the ln of a Negative NumberDate: 07/22/97 at 12:32:32 From: Cat Fyle Subject: Modern Algebra I have a test tomorrow in Modern Algebra, but the test is going to consist of complex numbers. I have not had complex numbers, so I have no idea where to begin. One of the questions is going to have something to do with finding the ln of a negative number. Can you give me a general formula for this problem? Like ln(-3)? Thank you very much, CAT Date: 07/22/97 at 12:58:26 From: Doctor Rob Subject: Re: Modern Algebra ln(x) = y means x = e^y. For what y is -3 = e^y? You need to know the formula e^(a+i*b) = e^a*[cos(b) + i*sin(b)] for real a and b. Then e^a = 3 and cos(b) + i*sin(b) = -1 leads to a = ln(3) and b = Pi*(2*k+1) for any integer k. Thus y = ln(-3) = ln(3) + i*Pi*(2*k+1) for any integer k. There is not one answer, but many (as with arctangent and other "functions" you have seen before). -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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