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Using Euler's relation, we know that e^(i*n*pi) = cos(n*pi) + i*sin(n*pi) where n is an integer. We also know that we can rewrite 10 as e raised to a specific power, namely e^(ln(10)). So substituting this back into 10^i and then applying Euler's relation, we obtain 10^i = (e^(ln(10)))^i = (e^(i*ln(10))) = cos(ln(10)) + i*sin(ln(10)).

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