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प्रश्न
If z1 = 2 – i and z2 = – 4 + 3i, find the inverse of z1, z2 and `("z"_1)/("z"_2)`
उत्तर
z1 = 2 – i, z2 = – 4 + 3i
z1z2 = (2 – i)(– 4 + 3i)
= – 8 + 3 + 4i + 6i
= – 5 + 10i
Inverse of z1z2 = (z1z2)–1
= `1/("z"_1"z"_2)`
= `1/(- 5 + 10"i") xx (- 5 - 10"i")/(- 5 - 10"i")`
= `(- 5 - 10"i")/((- 5)^2 - (10"i")^2`
= `(- 5 - 10"i")/(25 + 100)`
= `(-5(1 + 2"i"))/125`
= `(-(1 + 2"i"))/25`
`("z"_1)/("z"_2) = (2 - "i")/(-4 + 3"i") xx (- 4 - 3"i")/(-4 - 3"i")`
= `(- 8 - 6"i" + 4"i" + 3"i"^2)/((- 4)^2 - (3"i")^2`
= `(- 8 - 2"i" - 3)/(16 + 9)`
= `(- 11 - 2"i")/25`
= `(-(11 + 2"i"))/25`
Inverse of `("z"_1)/("z"_2) = (("z"_1)/("z"_2))^-1`
= `("z"_2)/("z"_1)`
= `(- 25)/(11 + 2"i") xx (11 - 2"i")/(11 - 2"i")`
= `(- 25(11 - 2"i"))/((11)^2 - (2"i")^2`
= `(- 25(11 - 2"i"))/(121 + 4)`
= `(- 25(11 - 2"i"))/125`
= `(-(11 - 2"i"))/5`
= `(- 11 + 2"i")/5`
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