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What are the differences between e^x and e^ix both graphically and theoretically?And my second question is what is the precise intuition behind rotation of 2D number and exponential function?

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Hint: $e^{ix}=\cos x + i\sin x$. This is called the Euler's formula and sometimes written as $\text{cis } x$.

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Well, if you're allowing complex valued $x$ there isn't a whole lot of theoretical difference. Just have $f(x)\to f(ix).$ One is just a "rescaled" version of the other (I use quotes cause in the complex plane "rescaling" x by $i$ is the same as rotating by 90 degrees. See more on this below.).

However if you're restricting to a real-valued $x$ then there's a world of difference. $e^x$ is a real valued function that you can graph simply. Famously, it is a positive, concave up curve that increases exponentially as $x\to \infty$ and asymptotes to zero as $x\to-\infty.$

On the other hand $e^{ix}$ can be written $\cos(x)+i\sin(x),$ so is a complex-valued function. If you watch the value on the complex plane as you increase $x$ you will see it wind and wind around the unit circle in a counterclockwise fashion. This is related to the parametrization of the unit circle $(x(t),y(t)) = (\cos(t),\sin(t)).$

If you multiply a complex number $z$ by $e^{ix},$ you can understand what's happening by writing the complex number in polar coordinates: $z=re^{i\theta} = r\cos(\theta) + ir\sin(\theta).$ Then we have $$ ze^{ix} = re^{i\theta}e^{ix} = re^{i(\theta+x)}$$ (To see this you can either trust the rules of exponentiation work in this complex territory or write everything out as sines and cosines and use trig identities.)

Thus the effect of multiplying by $e^{ix}$ for $x$ real is to add $x$ to the polar angle coordinate of the number, or, in simpler terms to rotate it counterclockwise by angle $x$.

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    But as every real number x is complex number x+0i then,should we realise e^(ix+a) is represent a concave upward curve in a special situation when x=0?if so then how can it be possible to transform the same function a circle to concave up curve? – 2017-02-19
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    @Abusayed Well the circle and the curve for the respective situations are actually different things. When you plot $e^x$ as a concave up curve, the x value is $x$ and the y axis is $e^x$.. it's the graph of a function in the usual sense. Whereas for $f(x) = e^{ix}$ the circle is the image of the function in the complex plane. The x value is the real part of $f$ and the y value is the imaginary part of $f.$ So it's not a graph in the same sense. To 'graph' a function from the reals to the complex, you would need 3 dimensions, one for $x$, two for real and imaginary part of f(x). It's a helix. – 2017-02-19
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    @Abusayed It's best to think of $f(z) = e^z$ as a function from the complex numbers to the complex numbers. (Which would require 4 dimensions to 'graph', so it's better to visualize in other ways). As you take move $z$ out in the imaginary direction the image winds around in a circle and as you take it out in the real direction, the distance from the origin in the complex plane increases exponentially. – 2017-02-19
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    Please explain me what I don't understand is that why we would require 4 dimension to plot the graph of f(z)=e^z where there r only 2 independent and one dependent variable? – 2017-03-03
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    @Abusayed The dependent variable is a complex number so it has two axes. One for real part and one for imaginary part. – 2017-03-03
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Multiplication by $e^{i\theta}$ (where $\theta$ is real) corresponds to a rotation of $\theta$ about the origin. This is best checked by using the polar form $z=r\cdot e^{i\theta}$. Perhaps a more basic approach is using Euler's formula

$$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$

and trigonometry (angle sum formulas) to justify why multiplication in polar form works the way it does. If you want to go even deeper, however, you will probably need to use power series to justify Euler's formula itself.

Now, since

$$i=\exp\left({\displaystyle i\frac{\pi}{2}}\right),$$

we have that $ix$ is obtained from $x$ by a $90^\circ$ degree rotation. The maps $e^{ix}$ and $e^x$ are hence related by this rotation.

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    But what about my first question -How could I realise the difference between e^x and e^i0+x graphically? – 2017-02-19