3i^30-i^19/2i-1

Question

- Change the question to general form, a+bi

Answer 1

3i^30-i^19/2i-1

Change the question to general form, a+bi

remember i has only four values

i = sqrt(-1), i^2 = -1, i^3 =-sqrt( -i), i^4 = -1

because we can divide multiples of i^4 they become values of 1

so look at the exponents 30 divided by 4 = 7 with remainder 2:  therefore we can cange the first exponent to 2

19 divided by 4 = 4 remained 3 this one to 3

rewrite as 3i^2 - i^3/2i - 1 now multiply the top and bottom by the conjugate 2i + 1

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

6i^3 - 2i^4 + 3i^2 - i^3 / 4i^2 + 2i - 2i - 1  =     now plug in again use the four values

-2i^4 + 5i^3 + 3i^2 / 4 i^2 - 1 = -2(1) + 5i^3 + 3(-1) / 4 (-1) - 1

5 i^3 -5 / -5 = -i^3 + 1

Answer 2

Given (3i^30-i^19)/(2i-1)
(3i^30)/(2i-1)-i^19/(2i-1)
(0.6+1.2i)-(-0.4+0.2i)
1+i

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Answer 3

(3i^30-i^19)/(2i-1) ··· Eq.1

Here, i^2=-1, i^3=(i^2-i)xi=-i, i^4=(i^2)x(i^2)=1, so i^30=(i^2)^15=-1, i^19=i^(16+3)={(i^4)^4}xi^3=-i

Plug i^30=-1, and i^19=-i into Eq.1.   Eq.1 restated as follows:

(3i^30-i^19)/(2i-1)={3(-1)+i}/(2i-1)=(i-3)/(2i-1) ··· Eq.2

To rationalize the denominator of Eq.2, multiply both numerator and denominator by the complex conjugate (2i+1).   Eq.2 restated as follows:

(i-3)/(2i-1)=(i-3)(2i+1)/(2i-1)(2i+1)={2(i^2)-5i-3}/{4(i^2)-1}={2(-1)-5i-3}/{4(-1)-1}={-5(i+1)}/-5 =i+1 

The answer is: (3i^30-i^19)/(2i-1)=i+1