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

The solution to ((2+i)(3-2i)(1+2i))/((1-i)^2) is (10+15i)/((1-i)^2)

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

simplify

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

\frac{10 + 15 i}{( 1 - i ) ^{2}}

Solution steps

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

Apply complex arithmetic rule:

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

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

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

Apply complex arithmetic rule:

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

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

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

Simplify

((2 \cdot 3 - 1 \cdot (- 2)) \cdot 1 - (2 (- 2) + 1 \cdot 3) \cdot 2) + ((2 \cdot 3 - 1 \cdot (- 2)) \cdot 2 + (2 (- 2) + 1 \cdot 3) \cdot 1) i : 10 + 15 i

= \frac{10 + 15 i}{( 1 - i ) ^{2}}

V = \frac{1}{3} π^{2}

(\frac{4 a ^{3}}{2 b ^{4}})^{2}

Q = 1600 - 20 \cdot 32

x + 11 > 16

(2 x + 1) (4 - 3 x) = 0

What is ((2+i)(3-2i)(1+2i))/((1-i)^2) ?
The solution to ((2+i)(3-2i)(1+2i))/((1-i)^2) is (10+15i)/((1-i)^2)