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

The solution to ((1+i)(1-2i))/((2+i)(4-3i)) is 7/25-1/25 i

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

simplify

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

\frac{7}{25} - \frac{1}{25} i

Solution steps

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

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

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

(2 + i) (4 - 3 i) = 11 - 2 i

= \frac{3 - i}{11 - 2 i}

Multiply by the conjugate

\frac{11 + 2 i}{11 + 2 i}

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

Simplify

\frac{( 3 - i ) ( 11 + 2 i )}{( 11 - 2 i ) ( 11 + 2 i )} : \frac{35 - 5 i}{125}

= \frac{35 - 5 i}{125}

Apply the fraction rule:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

= \frac{35}{125} + \frac{- 5}{125} i

Simplify

\frac{35}{125} + \frac{- 5}{125} i : \frac{7}{25} - \frac{1}{25} i

= \frac{7}{25} - \frac{1}{25} i

2 x^{2} + 24 x + 70

a^{3} - 2 a^{2} b + 3 a^{2} - 6 ab

2^{x} \geq 2500

∣ 3 x + 4 ∣ = 14

a^{2} - a - 1 = 0

What is ((1+i)(1-2i))/((2+i)(4-3i)) ?
The solution to ((1+i)(1-2i))/((2+i)(4-3i)) is 7/25-1/25 i