What is i(1-i)(2-i)(2+6i) ?

The solution to i(1-i)(2-i)(2+6i) is 20i

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simplify i(1-i)(2-i)(2+6i)

simplify

i (1 - i) (2 - i) (2 + 6 i)

20 i

Solution steps

i (1 - i) (2 - i) (2 + 6 i)

Apply complex arithmetic rule:

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

= i ((1 \cdot 2 - (- 1) (- 1)) + (1 \cdot (- 1) - 1 \cdot 2) i) (2 + 6 i)

Apply complex arithmetic rule:

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

= i (((1 \cdot 2 - (- 1) (- 1)) \cdot 2 - (1 \cdot (- 1) - 1 \cdot 2) \cdot 6) + ((1 \cdot 2 - (- 1) (- 1)) \cdot 6 + (1 \cdot (- 1) - 1 \cdot 2) \cdot 2) i)

Simplify

((1 \cdot 2 - (- 1) (- 1)) \cdot 2 - (1 \cdot (- 1) - 1 \cdot 2) \cdot 6) + ((1 \cdot 2 - (- 1) (- 1)) \cdot 6 + (1 \cdot (- 1) - 1 \cdot 2) \cdot 2) i : 20

= i 20

Rewrite in standard complex form:

20 i

= 20 i

- 6 x + x

- n + 4 (n + 1)

(r + 3)^{3}

(3 + 2 i) \cdot (4 + 5 i)

\frac{a ^{4} c ^{6}}{a ^{2} c}