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i^38 written in the standard form a+bi?
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\[i^{38} = -1\]
i,-1,-i,1
this cycle
I need an expression written in the a+bi form of i^38
\[-1 + 0 \times i\]
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thanks I'll try that
sure : )
ok -1 worked, using mathXL and don't like it
\[i^2 = -1\]\[\large \implies i^4 = i^2\cdot i^2 = (-1)(-1) = 1\]\[\large \implies i^{38} = \left(i^4\right)^9\cdot i^2 = 1^9(-1) = 1(-1) = -1\]
or think like this:4 goes in to 38 leaves a remainder of 2. so \[I^{38}=i^2=-1\] standard form is \[-1+0i\]
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