1. ## Complex number question

Is there a relatively fast way of solving this?

If $f(i)=\frac{1+i+i^2+...+i^{11}}{1-i}$ which of the following statements are true?

(a) $f(i)=2+i$

(b) $Re[f(i)]=5$

(c) $Im[f(i)]=8$

(d) $f(i)=1-i$

(e) $f(i)=0$

2. You can simplify the equation by using the fact that :

i^2 = -1
i^3 = (i)(i^2)=-i
i^4 = (i^2)^2 = 1
i^5 = (i^4) (i)=i
and so on

3. You can multiply the f(i) by 1+i/1+i to get rid of the complex number in the denominator..

4. Yep. But I was wondering if there's a faster way to do it.

5. How about muliplying by i-1?

6. songoku's method is the fastest. You need to recognize the fact that the powers of i cycle through four different numbers: 1, i, -1, and -i. The sum of any four consecutive powers of i will be 0. The fraction could be written like this:

$\frac{i^0 + i^1 + i^2 + i^3 + ... + i^{11}}{i - 1}$

The sum of the first four powers of i shown above is

$i^0 + i^1 + i^2 + i^3 = 1 + i - 1 - i = 0$

So how many powers of i are being added all together? What is the sum in the numerator? If you can answer these questions, you'll know what the fraction simplifies to.

01