What is (6+3i)+(8+4i)-(1+4i) ?

The solution to (6+3i)+(8+4i)-(1+4i) is 13+3i

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simplify (6+3i)+(8+4i)-(1+4i)

simplify

(6 + 3 i) + (8 + 4 i) - (1 + 4 i)

13 + 3 i

Solution steps

(6 + 3 i) + (8 + 4 i) - (1 + 4 i)

Simplify

(6 + 3 i) + (8 + 4 i) - (1 + 4 i) : 6 + 3 i + 8 + 4 i - (1 + 4 i)

= 6 + 3 i + 8 + 4 i - (1 + 4 i)

Group like terms

= 3 i + 4 i - (1 + 4 i) + 6 + 8

Add similar elements:

3 i + 4 i = 7 i

= 7 i - (1 + 4 i) + 6 + 8

Add the numbers:

6 + 8 = 14

= 7 i - (1 + 4 i) + 14

Remove parentheses:

- (a + bi) = - a - bi

= 7 i + - 1 - 4 i + 14

Group like terms

= 7 i - 4 i - 1 + 14

Add similar elements:

7 i - 4 i = 3 i

= 3 i - 1 + 14

Add the numbers:

- 1 + 14 = 13

= 3 i + 13

Rewrite in standard complex form:

13 + 3 i

= 13 + 3 i

(x^{2} p q^{2})^{2} (x p^{2})^{- 1}

\frac{3 ^{6}}{3 ^{4}}

- 3 x^{2} - 4 x = 0

\frac{6 x ^{- 2} y ^{2}}{3 y ^{- 4}}

6 x^{2} - 2 x = x