What is the mistake in the following process? #e^(pii)=-1# Let #a# and #b# real numbers. #(e^(cancelpii))^(b/cancelpi)=(-1)^(b/pi)# #e^(bi+a)=(-1)^(b/pi)*e^(a)# Natural log. on both sides. #a+bi=ln[(-1)^(b/pi)*e^(a)]# Now...

Let #a+bi=3+4pii#

#3+4pii=ln[(-1)^((4pi)/pi)*e^(3)]#

#3+4pii=ln[1*e^(3)]#

#3+4pii=3#

1 Answer
NJ
Mar 29, 2018

See comments for more discussion of this.

Explanation:

In general, for a complex number #z# of the form #z = a+bi#:

#=> ln(e^z) ne z#

The actual property is:

#=> color(blue)(ln(e^z) = z + 2pi i k)#

where #k# is an arbitrary integer.

In the case where #k = 0#, then it actually does hold that:

#=>ln(e^z) = z#

However, only considering #a# and #b# to be real numbers does not specify this one solution. Hence, as Douglas pointed out in the comments, we consider all solutions and only even multiples of #pi# are valid solutions for the complex logarithm. So in general, we cannot go from

#=> e^(bi+a) = (-1)^(b/pi)*e^a#

to

#=> a+bi = ln[(-1)^(b/pi)*e^a]#

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