What is (1+i)(a-bi)(3+i) ?

The solution to (1+i)(a-bi)(3+i) is (2a+4b)+(4a-2b)i

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simplify (1+i)(a-bi)(3+i)

simplify

(1 + i) (a - bi) (3 + i)

(2 a + 4 b) + (4 a - 2 b) i

Solution steps

(1 + i) (a - bi) (3 + i)

Apply complex arithmetic rule:

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

= ((1 \cdot a - 1 \cdot (- b)) + (1 \cdot (- b) + 1 \cdot a) i) (3 + i)

Apply complex arithmetic rule:

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

= ((1 \cdot a - 1 \cdot (- b)) \cdot 3 - (1 \cdot (- b) + 1 \cdot a) \cdot 1) + ((1 \cdot a - 1 \cdot (- b)) \cdot 1 + (1 \cdot (- b) + 1 \cdot a) \cdot 3) i

Simplify

((1 \cdot a - 1 \cdot (- b)) \cdot 3 - (1 \cdot (- b) + 1 \cdot a) \cdot 1) + ((1 \cdot a - 1 \cdot (- b)) \cdot 1 + (1 \cdot (- b) + 1 \cdot a) \cdot 3) i : (2 a + 4 b) + (4 a - 2 b) i

= (2 a + 4 b) + (4 a - 2 b) i

\frac{( x ^{2} - x ) ( x + 1 )}{x - 1}

(x - \frac{15}{4})^{2}

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

a [x^{2} + (- r - s) x + rs]

x sin (\frac{πn x}{L}) (L - x)