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mathsComplex NumbersAlgebra of Complex Numbers

Division of complex numbers

Complex numbers division is explained.

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Consider two complex numbers `z_1=a_1+ib_1` and `z_2 = a_2+ib_2`. What is `z_1 / z_2` in `a+bi` form?

  • `(a_1a_2+b_1b_2) + i (a_2b_1-a_1b_2)`
  • `a_1/a_2+ib_1/b_2`
  • `((a_1a_2+b_1b_2)+i(a_2b_1-a_1b_2))/(a_2^2+b_2^2)`
  • `((a_1a_2+b_1b_2)+i(a_2b_1-a_1b_2))/(a_2^2+b_2^2)`

The answer is '`((a_1a_2+b_1xxb_2)+i(a_2b_1-a_1b_2))/(a_2^2+b_2^2)`'

`z_1=a_1+ib_1` and `z_2 = a_2+ib_2`. What is `z_1 / z_2`?
Solution :
`z_3 = z_1 / z_2 `
`quad quad = (a_1+ib_1)/(a_2+ib_2)`
`quad quad = (a_1+ib_1)/(a_2+ib_2) xx (a_2-ib_2)/(a_2-ib_2) `
`quad quad = ((a_1+ib_1)xx (a_2-ib_2))/((a_2+ib_2) xx (a_2-ib_2)) `
`quad quad = ((a_1a_2+b_1b_2)+i(a_2b_1-a_1b_2))/(a_2^2+b_2^2)`

Division of two complex numbers follows numerical expression laws and properties with `i` handled as `(sqrt(-1))^2 = -1` to arrive at `a+bi` form.

Division of two complex numbers : For any complex number `z_1=a_1+ib_1 in CC` and `z_2 = a_2+ib_2 in CC`
`z_1 / z_2 =` `((a_1a_2+b_1b_2)+i(a_2b_1-a_1b_2))/(a_2^2+b_2^2)`

Solved Exercise Problem:

Given `z_1 = 10+5i` and `z_2=3-4i` what is `z_1/z_2`?

  • `(6+8i)/5`
  • `(3i+4)/25`
  • `10+11i`
  • `(10+11i)/5`
  • `(10+11i)/5`

The answer is '`(10+11i)/5`'.

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