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OpenStudy (anonymous):

i^38 written in the standard form a+bi?

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OpenStudy (anonymous):

\[i^{38} = -1\]

OpenStudy (anonymous):

i,-1,-i,1

OpenStudy (anonymous):

this cycle

OpenStudy (anonymous):

I need an expression written in the a+bi form of i^38

OpenStudy (anonymous):

\[-1 + 0 \times i\]

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OpenStudy (anonymous):

thanks I'll try that

OpenStudy (anonymous):

sure : )

OpenStudy (anonymous):

ok -1 worked, using mathXL and don't like it

OpenStudy (anonymous):

\[i^2 = -1\]\[\large \implies i^4 = i^2\cdot i^2 = (-1)(-1) = 1\]\[\large \implies i^{38} = \left(i^4\right)^9\cdot i^2 = 1^9(-1) = 1(-1) = -1\]

OpenStudy (anonymous):

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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