(1+(2*I))/(3-(4*I))+2/5=(I-2)/(5*I)

Answer 1

To solve this equation, we first simplify the left-hand side by finding a common denominator for the two fractions:

(1 + 2i)/(3 - 4i) + 2/5
= (1 + 2i)/(3 - 4i) + (2/5) * ((3 + 4i)/(3 + 4i)) // multiplying the second fraction by 1 in the form of (3+4i)/(3+4i)
= (1 + 2i)/(3 - 4i) + (6 + 8i)/(15 + 20i)
= [(1 + 2i)*(15 + 20i) + (6 + 8i)*(3 - 4i)] / [(3 - 4i)*(15 + 20i)]
= (-46 - 7i)/221

Now we simplify the right-hand side by multiplying both the numerator and denominator by -i to eliminate the imaginary numbers in the denominator:

(i - 2)/(5i) = [(i - 2)*(-i)] / (5i * -i) = (2 + i)/5

So our original equation becomes:

(-46 - 7i)/221 = (2 + i)/5

Multiplying both sides by 221, we get:

-46 - 7i = 221*(2 + i)/5

Multiplying both sides by 5 and expanding the right-hand side, we get:

-230 - 35i = 442 + 221i

Simplifying, we get:

186 = 256i

Dividing both sides by 256, we get:

i = 186/256 = 93/128

So the solution to the equation is:

i = 93/128

Answer 2

AAAaannndd the bot gets it wrong yet again!

what rubbish! You can't use i as the imaginary unit, and then solve to show that i = 93/128