25i/(3+4i)=3i

Solution for 25i/(3+4i)=3i equation:

25i/(3+4i)=3i

We move all terms to the left:

25i/(3+4i)-(3i)=0


Domain of the equation: (3+4i)!=0

We move all terms containing i to the left, all other terms to the right

4i!=-3

i!=-3/4

i!=-3/4

i∈R


We add all the numbers together, and all the variables

25i/(4i+3)-3i=0

We add all the numbers together, and all the variables

-3i+25i/(4i+3)=0

We multiply all the terms by the denominator


-3i*(4i+3)+25i=0

We add all the numbers together, and all the variables

25i-3i*(4i+3)=0

We multiply parentheses

-12i^2+25i-9i=0

We add all the numbers together, and all the variables

-12i^2+16i=0

a = -12; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-12)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-12}=\frac{-32}{-24}
=1+1/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-12}=\frac{0}{-24}
=0
$