What are exponents? Idea behind exponents(complex or real)? I recently came through an article which said that $e^{\iota x}$ means that a $e$ gradually increases every moment by a factor of $\iota x$ perpendicular to the real part which I took as a force of $\iota x$ is applied perpendicular to a string of length $e$ same as in circular motion but the expression $e^{\iota x}$ has radius $1$ instead of $e$ in complex plane.
I'm very confused with the idea behind exponents.
 A: Given an element $a$ of some multiplicative structure with identity element $1$ one defines recursively
$$a^0:=1,\qquad a^{n+1}:=a\cdot a^n\quad(n\geq0)\ .$$
The resulting expressions of the form $a^n$ are called powers. The variable $a$ is called the base, and the variable $n$ is called the exponent. A less sophisticated way to define $a^n$ is writing $a^n:=a\cdot a\cdot\ldots\cdot a$ ($n$ factors). I guess you knew this all along. 
So far the exponent had to be a natural number. One then extends the definition to integer and even to arbitrary rational exponents, whereby one assumes $a>0$ for simplicity. So far the idea "$a^n:=a\cdot a\cdot\ldots\cdot a$ ($n$ factors)" can be more or less sustained. In order to define  things like $2^\sqrt{2}$ however one has to resort to continuity, but o.k. In summa one can rescue the power laws
$$a^{x+y}=a^x\cdot a^y,\quad{\rm etc.}\tag{1}$$
all the way through.
Now in complex analysis one defines in a formal way the function $\exp( z):=\sum_{k\geq0}{z^k\over k!}$. One then proves that this function universally satisfies the functional equation $(1)$:
$$\exp(z+w)=\exp z\cdot\exp(w)\qquad\forall z,\>w\in{\mathbb C}\ .$$
This implies that for arbitrary integer or rational $r$ one has $\exp(r)=\bigl(\exp(1)\bigr)^r=e^r$, wereby $e$ is an abbreviation for $\exp(1)$. Instead of $\exp(z)$ one then writes $e^z$ for arbitrary complex $z$, even though an interpretation as repeated multiplication is impossible for such $z$.
The rest of my answer assumes that you have a minimal familiarity with complex numbers.
Assume that $x$is real with $0<x\ll1$. Then the quantity
$$e^{ix}=1+ix-{x^2\over2!}-i{x^3\over3!}+{x^4\over4!}+\ldots\approx 1+ix$$
is on the unit circle  slightly above the point $1\in{\mathbb C}$. In fact the polar angle of $e^{ix}$ is given by ${\rm Arg}(e^{ix})=x$. 
If $a_0=a\ne0$ is any nonzero complex number then the operation
$$a_n\mapsto a_{n+1}=e^{ix}\>a_n\qquad(n\geq0)$$
rotates $a_n$ by the tiny angle $x$ around the origin to its new position $a_{n+1}$. One then has
$$a_n=\bigl(e^{ix}\bigr)^n \>a=e^{i\,nx}\>a\qquad(n\geq0)\ ,$$
which says that in $n$ steps we obtain a rotation by the angle $nx$.
