What is (3+2i)^2-(4+2i) ?

The solution to (3+2i)^2-(4+2i) is 1+10i

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simplify (3+2i)^2-(4+2i)

simplify

(3 + 2 i)^{2} - (4 + 2 i)

1 + 10 i

Solution steps

(3 + 2 i)^{2} - (4 + 2 i)

(3 + 2 i)^{2} = 12 i + 5

- (4 + 2 i) = - 4 - 2 i

= 12 i + 5 - 4 - 2 i

Group like terms

= 12 i - 2 i + 5 - 4

Add similar elements:

12 i - 2 i = 10 i

= 10 i + 5 - 4

Subtract the numbers:

5 - 4 = 1

= 10 i + 1

Rewrite in standard complex form:

1 + 10 i

= 1 + 10 i

225 + 169

(2 s + 1)^{3}

1 - 21

7 x + 5 - 4 x - 16

(2 (3) - i) (\frac{3}{5} - 2 i)