Question 70009
Recognize that by definition {{{i^2 = -1}}} and {{{i^1=i}}}

From this we can develop the sequence:

{{{i = i}}}
{{{i^2 = -1}}}
{{{i^3 = i^2 * i = (-1)*i = -i}}}
{{{i^4 = i^2*i^2 = (-1)*(-1) = 1}}}
{{{i^5 = i^4*i = 1*i = i}}}
{{{i^6 = i^3*i^3 = (-i)*(-i) = i^2 = -1}}}
{{{i^7 = i^6*i^1 = (-1)*i = -i}}}
{{{i^8 = i^4*i^4 = 1*1 = 1}}}
{{{i^9 = i^8*i^1 = 1*i = i}}}

Note that this series is i, -1, -i, 1, i, -1, -i, 1, i, ....

We can use this table along with some rules of exponents to simplify the terms in the problem.

Let's try to simplify {{{i^90}}}.  We use one of the laws of exponents to rewrite
the this as {{{i^90 = (i^9)^10}}}.  From the above table you can see that {{{i^9 = i}}}
Substitute this to get {{{i^90 = (i)^10}}}.  We can further use a law of exponents
to factor this term into {{{i^10 = i^5*i^5}}}.  But {{{i^5 = i}}} so:
{{{i^5*i^5=i*i=i^2 = -1}}} 
So {{{i^90=-1}}}.

We can use this result to find {{{i^91}}}. Note that {{{i^91 = i^90*i^1}}}.  Substitute
-1 for {{{i^90}}} to get {{{i^91 = (-1)*i = -i}}}.

Then use this result and procedure to find that: 
{{{i^92 = i^91*i^1 =(-i)*i = -i^2 = -(-1) = 1}}}

Repeat this process to find that:
{{{i^93 = i^92*i^1 = 1*i = i}}}

Substitute these four results into the original problem:
{{{(i^90)*(i^91) + (i^92)*(i^93)}}}

The substitution results in:

{{{(-1)*(-i) + (1)*(i) = i + i = 2i}}}

The answer to your problem is 2i.

Hope this helps.