Mathematics 95 Online
OpenStudy (anonymous):

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

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):

$-1 + 0 \times i$

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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