For an assignment in one of my math classes I have this problem. Here is where I have gotten so far. This is in V, an inner product space over $\mathbb{C}$
$$\frac{1}{4} \sum^4_{k=1}i^k \left \Vert u+i^kv \right \Vert = \frac{1}{4} \sum^4_{k=1} i^k \langle u+i^kv, u+i^kv \rangle\\ = \frac{1}{4} \sum^4_{k=1} i^k \langle u, u+i^kv \rangle + i^k \langle i^kv, u+i^kv\rangle \\ =\frac{1}{4} \sum^4_{k=1} i^k\langle u,u+i^kv \rangle + i^{2k} \langle v, u+i^kv\rangle\\ =\frac{1}{4} \sum^4_{k=1} i^k \overline{\langle u+i^kv, u \rangle} + i^{2k} \overline{\langle u+i^kv, v\rangle}\\ =\frac{1}{4} \sum^4_{k=1} i^k \overline{\langle u, u \rangle}+i^k \overline{\langle i^kv, u \rangle} + i^{2k} \overline{\langle u,v\rangle}+i^{2k} \overline{\langle i^kv,v\rangle}\\ =\frac{1}{4} \sum^4_{k=1} i^k \overline{\langle u, u \rangle}+i^k\overline{i^k} \overline{\langle v, u \rangle} + i^{2k} \overline{\langle u,v\rangle}+i^{2k}\overline{i^k} \overline{\langle v,v\rangle}\\ =\frac{1}{4} \sum^4_{k=1} i^k \langle u, u \rangle-i^{2k} \langle u, v \rangle + i^{2k} \langle v,u\rangle-i^{3k} \langle v,v\rangle$$ After this, I try expanding the summation but it results in a very complicated series of sums and it does not reduce do $\langle u,v \rangle$. Any mistakes I did, or tips would be appreciated!
EDIT Found the solution! I should've stopped at step 3 $\frac{1}{4} \sum^4_{k=1} i^k\langle u,u+i^kv \rangle + i^{2k} \langle v, u+i^kv\rangle$. Here you just do the sum for each value of k (k=1,2,3,4) and you will be able to eliminate every term and get $4\langle u,v \rangle$ .
A somewhat simpler approach uses $$\langle u+i^k v,\,u+i^k v\rangle=\langle u,\,u\rangle+(-1)^k\langle v,\,v\rangle+(i^{-k}\langle u,\,v\rangle+\text{c.c.})$$(note the plus-complex-conjugate grouping of cross terms). Your expression is then $$\frac{\sum_k \langle u,\,v\rangle+\sum_k i^k\langle u,\,u\rangle+\sum_k i^{2k}\langle v,\,u\rangle+\sum_k i^{3k}\langle v,\,v\rangle}{4}.$$Now use $\sum_k i^{jk}=4\delta_{j0}$ for $j\in\{0,\,\cdots,\,3\}$.