Write each number in standard form (a+ib). i^-1 13i(2+3i)/3+2i

For i⁻¹  : since a⁻ᵇ = 1/aᵇ, we have    the denominator is imake the denominator real by multiplying the numerator and denominator by i so that we get i² = -1    i⁻¹ in standard form is 0 - i======my second answer will assume that 3 + 2i is the entire denominator.    distribute 13i into (2+3i) first to get 26i + 39i², which turns into 26i + 39(-1), which turns into -39 + 26i.    make the denominator real by multiplying the numerator and denominator by the conjugate of the denominator 3 + 2i. the conjugate is 3 - 2i.note the the numerator becomes, via foiling or distributing (i use foil here)    (-39 + 26i)(3 - 2i)    = (-39)(3) + (-) + (26i)(3) + ()      = -117 + 78i + 78i - 52i²    = -117 + 156i - 52(-1)    = -117 + 156i + 52    = -65 + 156ithe denominator becomes    (3+2i)(3-2i)    = 3² + () + (2i)(3) - (2i)(2i)    = 3² - (2i)²    = 9 - 4i²    = 9 - 4(-1)    = 9 + 4 = 13so it follows that    13i(2+3i)/(3+2i) in standard form is -5 + 12i