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Thought-Process to Discover Knowledge

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mathsComplex NumbersAlgebra of Complex Numbers

Exponent of a complex Number

Exponent of a complex number is explained.



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Given `z_1 = a_1+ib_1` and `z_2 = a_2+ib_2`, Can the exponent `z_1^z_2` be written in `a+ib` form?

  • No. Exponent to the power of complex number is not possible
  • Yes. By converting to polar form `z_1 = r e^(i theta)`.
  • Yes. By converting to polar form `z_1 = r e^(i theta)`.

The answer is 'Yes. By converting to polar form `z_1 = r e^(i theta)'

Given `z_1 = a_1+ib_1` and `z_2 = a_2+ib_2`, exponent
`z_1^z_2`
`quad quad = (r_1e^(i theta_1) )^(a_2+ib_2)`
`quad quad = r_1^a_2 `
`quad quad quad quad xx r_1^(ib_2)`
`quad quad quad quad xx e^(i theta_1 a_2)`
`quad quad quad quad xx e^(i theta_1 i b_2)`

`quad quad = r_1^a_2 `
`quad quad quad quad xx e^(ib_2 ln r_1) (`
`quad quad quad quad xx e^(i theta_1 a_2)`
`quad quad quad quad xx e^(- theta_1 b_2)`

`quad quad = r_1^a_2 e^(- theta_1 b_2) `
`quad quad quad quad xx e^(i(b_2 ln r_1 + theta_1 a_2))`

The result is in the polar form and can be converted to coordinate form.

Given `z_1 = a_1+ib_1` and `z_2 = a_2+ib_2`, Can the root `root(z_2)(z_1)` be computed in `a+ib` form?

  • No. root to a complex number is not possible
  • Yes. By considering root as power of `(1/z_2)`
  • Yes. By considering root as power of `(1/z_2)`

The answer is 'Yes. By considering root as power of `(1/z_2)`'

Given `z_1 = a_1+ib_1` and `z_2 = a_2+ib_2`, root
`root(z_2)(z_1)`
`quad quad = z_1^(1/z_2)`
`quad quad = z_1^(bar(z_2)/(|z_2|^2))`
By following the rules of exponent of a complex number, the root can be solved.

To find exponent and root of complex numbers, the rules of numerical expression is used to arrive at the coordinate form `a+ib`.

Exponent and Roots of Complex number
 •  For `z_1^(z_2)`, convert `z_1` to polar form `re^(i theta)`
 •  For `a^(ib)`, convert `a` to `e^(ln a)` form
 •  For `z_1^(1/z_2)`, convert `1/z_2` to a complex number in numerator `bar(z_2)/(|z_2|^2)`

Solved Exercise Problem:

What is `(1+i)^(3/2)`

  • `3/2+3/2i`
  • `2^(3/4)(cos (3 pi/8)+i sin (3 pi/8))`
  • `2^(3/4)(cos (3 pi/8)+i sin (3 pi/8))`
  • `2^(3/2)(cos (3 pi/4)+i sin (3 pi/4))`
  • `2^(3/4)(cos (3 pi/2)+i sin (3 pi/2))`

The answer is '`2^(3/4)(cos (3 pi/8)+i sin (3 pi/8))`'

                            
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