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Note the repeating pattern

i^1  = i

i^2  = -1

i^3  =  -i

i^4  = 1

So....divide 1234 by 4  and we have

308. 5  =  308 + 1/2    =   308 + 2/4

The "2"  tells us that  i^1234 is equivalent to  i^2  =   -1

Compute \(\mathbf{i^{1234}}\).

\(\begin{array}{|rcll|} \hline && \mathbf{i^{1234}} \\ &=& \left(i^2\right)^{617} \quad | \quad i^2 = -1 \\ &=& \left(-1\right)^{617} \\ &=& \mathbf{ -1} \\ \hline \end{array} \)