How do I show this without using Euler Formula :$e^{i\frac{\pi}{2}}-i=e^{i\pi}+1$? I would like to show  this without using Euler Formula:$$e^{i\frac{\pi}{2}}-i=e^{i\pi}+1$$.?
Note: I multiplied both side by $i$ but i don't succeed .
Edit: I edited the question because i meant Euler formula and the last very related to the titled question
Thank you for any help 
 A: Euler's Identity refers to $e^{i \pi} + 1 = 0$, whereas Euler's formula refers to $e^{ix} = \cos(x) + i \sin(x)$. If it is just the identity that needs not to be used, then simply use Euler's formula and plug in the corresponding value for $x$.
On the to hand, if both are not allowed, then a quick derivation of Euler's formula using the sums of trigonometric functions, or the polar representation $e^{ix} = r\left(\cos(\theta) + i\sin(\theta)\right)$ should be sufficient.
A: Without defining what $e^z, z \in \mathbb{C}$ is, none of the $e^{i\pi},e^{i\frac{\pi}{2}}$ make sense. And, inevitably, this leads to Euler's formula. Now, depending on the complexity of the course, either 


*

*easy - define $e^{i\phi}=\cos{\phi}+i\sin{\phi}$, like this one, page 7, with some reasoning why this definition works, or

*more dificult - define $e^z$ using power series and prove the relation. 


Also, using geometry and vector representations of complex numbers:


*

*$e^{i\pi}$ is the exponential (and polar) form of the vector $(-1, 0)$. Now $(-1, 0) + (1, 0) = (0,0)$.

*$e^{i\frac{\pi}{2}}$ is the exponential (and polar) form of the vector $(0, 1)$ and $i=(0,1)$, so $(0, 1)-(0, 1)=(0,0)$.


One may ask why "inevitably", it's because power series are used to define even more "exotic" exponentials, like $e^A$, wher $A$ is a matrix.
