How do you solve #e^x=0#?

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Answer 1 · 2016-06-30T06:31:51

There is no #x# such that #e^x = 0#

The function #e^x# considered as a function of Real numbers has domain #(-oo, oo)# and range #(0, oo)#.

So it can only take strictly positive values.

When we consider #e^x# as a function of Complex numbers, then we find it has domain #CC# and range #CC "\" { 0 }#.

That is #0# is the only value that #e^x# cannot take.

Note that #e^(x+yi) = e^x e^(yi) = e^x(cos y+i sin y)#

We have already noted that iof #x in RR# then #e^x > 0#.

For pure imaginary exponents the result is on the unit circle, specifically:

#e^(yi) = cos y + i sin y != 0#

So #e^(x+yi) != 0# for all #x, y in RR#