What is |((1+i)^6)/((1-i)^2(sqrt(2)+i)^4)| ?

The solution to |((1+i)^6)/((1-i)^2(sqrt(2)+i)^4)| is |(4(-7-4sqrt(2)i))/(81)|

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simplify |((1+i)^6)/((1-i)^2(sqrt(2)+i)^4)|

simplify

\frac{( 1 + i ) ^{6}}{( 1 - i ) ^{2} ( 2 + i ) ^{4}}

\frac{4 ( - 7 - 4 2 i )}{81}

Solution steps

\frac{( 1 + i ) ^{6}}{( 1 - i ) ^{2} ( 2 + i ) ^{4}}

Expand

(1 + i)^{6} : - 8 i

= \frac{- 8 i}{( 1 - i ) ^{2} ( 2 + i ) ^{4}}

Simplify

(1 - i)^{2} (2 + i)^{4} : - 2 i (42 i - 7)

= \frac{- 8 i}{- 2 i ( 4 2 i - 7 )}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

\frac{- 8 i}{- 2 i ( 4 2 i - 7 )} = \frac{8 i}{2 i ( 4 2 i - 7 )}

= \frac{8 i}{2 i ( 4 2 i - 7 )}

Cancel the common factor:

i

= \frac{8}{2 ( 4 2 i - 7 )}

Factor the number:

8 = 2 \cdot 4

= \frac{2 \cdot 4}{2 ( 4 2 i - 7 )}

Cancel the common factor:

2

= \frac{4}{4 2 i - 7}

Simplify

\frac{4}{4 2 i - 7} : \frac{4 ( - 7 - 4 2 i )}{81}

= \frac{4 ( - 7 - 4 2 i )}{81}

2 (n (m + 1))

\frac{1}{( s + 1 ) \cdot ( s + 2 )}

(x^{\frac{1}{2}} + x^{\frac{2}{3}})^{9}

a (4, - 5) b (10, 4)

(x + 5) (x - 7)

What is |((1+i)^6)/((1-i)^2(sqrt(2)+i)^4)| ?
The solution to |((1+i)^6)/((1-i)^2(sqrt(2)+i)^4)| is |(4(-7-4sqrt(2)i))/(81)|