What is (1+i)(1-2i)(1+3i) ?

The solution to (1+i)(1-2i)(1+3i) is 6+8i

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simplify (1+i)(1-2i)(1+3i)

simplify

(1 + i) (1 - 2 i) (1 + 3 i)

6 + 8 i

Solution steps

(1 + i) (1 - 2 i) (1 + 3 i)

Apply complex arithmetic rule:

(a + bi) (c + d i) = (a c - b d) + (a d + b c) i

= ((1 \cdot 1 - 1 \cdot (- 2)) + (1 \cdot (- 2) + 1 \cdot 1) i) (1 + 3 i)

Apply complex arithmetic rule:

(a + bi) (c + d i) = (a c - b d) + (a d + b c) i

= ((1 \cdot 1 - 1 \cdot (- 2)) \cdot 1 - (1 \cdot (- 2) + 1 \cdot 1) \cdot 3) + ((1 \cdot 1 - 1 \cdot (- 2)) \cdot 3 + (1 \cdot (- 2) + 1 \cdot 1) \cdot 1) i

Simplify

((1 \cdot 1 - 1 \cdot (- 2)) \cdot 1 - (1 \cdot (- 2) + 1 \cdot 1) \cdot 3) + ((1 \cdot 1 - 1 \cdot (- 2)) \cdot 3 + (1 \cdot (- 2) + 1 \cdot 1) \cdot 1) i : 6 + 8 i

= 6 + 8 i

(9 x^{2} - 4) - (10 x^{2} + 9 x - 8)

5 h - 6 - 8 + 7 h

(x + 2)^{2} - 5 (x + 2)

\frac{- 16}{- 8}