Question

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

Answer 1

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

Answer 2

i,-1,-i,1

Answer 3

this cycle

Answer 4

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

Answer 5

\[-1 + 0 \times i\]

Answer 6

thanks I'll try that

Answer 7

sure : )

Answer 8

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

Answer 9

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

Answer 10

or think like this:4 goes in to 38 leaves a remainder of 2. so
\[I^{38}=i^2=-1\] standard form is
\[-1+0i\]