What is (3+2i)/i+(6-i)^2-3i^2 ?

The solution to (3+2i)/i+(6-i)^2-3i^2 is 40-15i

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simplify (3+2i)/i+(6-i)^2-3i^2

simplify

\frac{3 + 2 i}{i} + (6 - i)^{2} - 3 i^{2}

40 - 15 i

Solution steps

\frac{3 + 2 i}{i} + (6 - i)^{2} - 3 i^{2}

3 i^{2} = - 3

= \frac{3 + 2 i}{i} + (6 - i)^{2} - (- 3)

Apply rule

- (- a) = a

= \frac{3 + 2 i}{i} + (6 - i)^{2} + 3

(6 - i)^{2} : 35 - 12 i

= \frac{3 + 2 i}{i} + 35 - 12 i + 3

Add the numbers:

35 + 3 = 38

= 38 + \frac{3 + 2 i}{i} - 12 i

\frac{3 + 2 i}{i} = 2 - 3 i

= 38 + 2 - 3 i - 12 i

Rewrite

38 + 2 - 3 i - 12 i

in standard complex form:

40 - 15 i

= 40 - 15 i

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\frac{74}{5 - 7 i} + \frac{15 - 20 i}{1 - 2 i} - \frac{10 i}{2 - i}