What is ((1-i)(2+3i)(3+i)(2-2i))/((-1-i)^3(4-i)) ?

The solution to ((1-i)(2+3i)(3+i)(2-2i))/((-1-i)^3(4-i)) is 48/17+46/17 i

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simplify ((1-i)(2+3i)(3+i)(2-2i))/((-1-i)^3(4-i))

simplify

\frac{( 1 - i ) ( 2 + 3 i ) ( 3 + i ) ( 2 - 2 i )}{( - 1 - i ) ^{3} ( 4 - i )}

\frac{48}{17} + \frac{46}{17} i

Solution steps

\frac{( 1 - i ) ( 2 + 3 i ) ( 3 + i ) ( 2 - 2 i )}{( - 1 - i ) ^{3} ( 4 - i )}

(1 - i) (2 + 3 i) (3 + i) (2 - 2 i) = 44 - 12 i

= \frac{44 - 12 i}{( - 1 - i ) ^{3} ( 4 - i )}

Simplify

(- 1 - i)^{3} (4 - i) : - 10 i + 6

= \frac{44 - 12 i}{- 10 i + 6}

Factor

44 - 12 i : 4 (11 - 3 i)

= \frac{4 ( 11 - 3 i )}{- 10 i + 6}

Factor

- 10 i + 6 : 2 (- 5 i + 3)

= \frac{4 ( 11 - 3 i )}{2 ( - 5 i + 3 )}

Factor the number:

4 = 2 \cdot 2

= \frac{2 \cdot 2 ( 11 - 3 i )}{2 ( - 5 i + 3 )}

Cancel the common factor:

2

= \frac{2 ( 11 - 3 i )}{- 5 i + 3}

Simplify

\frac{2 ( 11 - 3 i )}{- 5 i + 3} : \frac{48}{17} + \frac{46}{17} i

= \frac{48}{17} + \frac{46}{17} i

2 x + 2 (x + 2) > 112

3 x^{4} + 2 x^{2} - 1

\frac{1}{\frac{3}{5}}

\frac{3}{5} o f^{270}

\frac{π}{3} + \frac{2 π}{3}

What is ((1-i)(2+3i)(3+i)(2-2i))/((-1-i)^3(4-i)) ?
The solution to ((1-i)(2+3i)(3+i)(2-2i))/((-1-i)^3(4-i)) is 48/17+46/17 i