Here we calculate the exponent of a Vector. The result depends on whether the dimensions commute or anticommute and whether the dimensions square to positive or negative. A summary of the results is given in the following table:
commutative  square to  result  example  derivation 

anticommute  all positive  exp(v) = cosh(v) + norm(v)*sinh(v)  Euclidean vector  see below 
anticommute  all negative  exp(v) = cos(v) + norm(v)*sin(v)  bivector  see below 
commute  two positive  exp(x + Dy) = cosh(v) + D sinh(v)  double number  double number pages 
commute  one positive, one negative  r e^{iθ} = r (cos(θ) + i sin(θ))  complex number  complex number pages 
where:
 v = √(v•v)) = magnitude scalar value
 norm(v) = v*(1/√(v•v)) = normalised (unit length) vector
To summarise and interpret the results, if the dimensions commute (as they do for complex numbers for example) then the result is a pure vector but, if the dimensions anticommute (as they do for vectors in euclidean space for example) then the result is a scalar plus a vector. If this idea of adding scalars and vectors is a problem then see the pages about clifford (geometric) algebra, what this is saying is that the exponent does not have a solution within vector space but it does have a solution in clifford algebra.
in order to derive this result we will try two approaches:
 polar form
 infinite series
However, before we start, it will be useful to review the formula for infinite series.
Infinite Series
I have not found a version of Euler's equation which applies to vectors so we need to calculate it ourselves.
The only method that I can think of is to calculate the exponent using the series:
e^{(v)} = 


Where:
 v = vector
 n= integer
 e = 2.71828
We have to be careful with vectors because they are not in general commutative. I think the above series applies but I'm not absolutely sure.
We now need to plug in a value for (v)^{n} which we have calculated on this page.
sin(x)  x  x^{3}/3! + x^{5}/5! ... +(1)^{r}x^{2r+1}/(2r+1)!  all values of x 
cos(x)  1  x^{2}/2! + x^{4}/4! ... +(1)^{r}x^{2r}/(2r)!  all values of x 
ln(1+x)  x  x^{2}/2! + x^{3}/3! ... +(1)^{r+1}x^{r}/(r)!  1 < x <= 1 
exp(x)  1 + x^{1}/1! + x^{2}/2! + x^{3}/3! ... + x^{r}/(r)!  all values of x 
exp(x)  1  x^{1}/1! + x^{2}/2!  x^{3}/3! ...  all values of x 
e  1 + ^{1}/1! + ^{2}/2! + ^{3}/3!  =2.718281828 
sinh(x)  x + x^{3}/3! + x^{5}/5! ... +x^{2r+1}/(2r+1)!  all values of x 
cosh(x)  1 + x^{2}/2! + x^{4}/4! ... +x^{2r}/(2r)!  all values of x 
Case 1: all dimensions square to positive
We start with the series:
exp(v) = 1 + v^{1}/1! + v^{2}/2! + v^{3}/3!+ v^{4}/4! + v^{5}/5! + …
So if all dimensions square to positive then:
v^{2} = v•v = positive scalar
so substituting gives:
exp(v) = 1 + v + v•v/2! + v•v*v/3!+ v•v*v•v/4! + v•v*v•v*v/5! + …
splitting up into real and vector parts gives:
exp(v) = 1 + v•v/2! + v•v*v•v/4! +…+ v*( 1+ v•v/3! + v•v*v•v/5! + …)
So we can see that it is a scalar + vector*another scalar
exp(v) = cosh(√(v•v)) + v*(1/√(v•v))*sinh(√(v•v))
Case 2: all dimensions square to negative
We start with the series:
exp(v) = 1 + v^{1}/1! + v^{2}/2! + v^{3}/3!+ v^{4}/4! + v^{5}/5! + …
So if all dimensions square to negative then:
v^{2} = v•v = negative scalar
so substituting gives:
exp(v) = 1 + v  v•v/2!  v•v*v/3!+ v•v*v•v/4! + v•v*v•v*v/5! + …
splitting up into real and vector parts gives:
exp(v) = 1  v•v/2! + v•v*v•v/4! +…+ v*( 1 v•v/3! + v•v*v•v/5! + …)
So we can see that it is a scalar + vector*another scalar
exp(v) = cos(√(v•v)) + v*(1/√(v•v))*sin(√(v•v))
Case 3: dimensions square to mixture of positive and negative
The difference between the above two cases depends on whether v^{2} = a positive or negative scalar.
So if we have a vector 'v' made up of the vectors x,y and z then to determine the nature of the exponent we calculate
s = v•v = x² + y ²+ z²
as follows:
√(v•v)  exp(v) 

if √(v•v) = +ve then:  exp(v) = cosh(√(v•v)) + v/(√(v•v))*sinh(√(v•v)) 
if √(v•v) = ve then:  exp(v) = cos(√(v•v)) + v/(√(v•v))*sin(√(v•v)) 
if √(v•v) = 0 then:  exp(v) = 1 + v 