SOLUTION: (1+2i)/(3-4i)+(2-i)/5i

Algebra.Com
Question 933635: (1+2i)/(3-4i)+(2-i)/5i


Found 2 solutions by MathLover1, Alan3354:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!




...common denominator is , so add numerators



....multiply both by to eliminate from denominator









........divide both by






Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
(1+2i)/(3-4i)+(2-i)/5i
---------
Rationalize the DEN's first
---
(1+2i)*(3+4i)/25 + (2-i)*i/(-5)
= (3 + 10i - 8)/25 + (2i+i)/(-5)
= (3 + 10i - 8)/25 - (1 + 2i)/5
= (-5 + 10i)/25 - (1 - 2i)/5
= (-1 + 2i)/5 - (1 + 2i)/5
= -2/5

RELATED QUESTIONS

Simplify 3-2i-4i^3 /... (answered by robertb)
(2-4i)/(2) (2+4i)/(-5+2i) (1-i)/(-4-6i)... (answered by jsmallt9,palanisamy)
I wanted you to check my answers on these two problems: 1) Problem: (2-4i)(2-i)... (answered by checkley77,gonzo)
(3-71)(3+7i) (2-2i)(2+2i) (7-10i)(7+10i) (2+5i)(2-5i) (4+3i)(1-i) (5-7i)(1+2i)... (answered by ewatrrr)
How do I solve complex number problems like 6+5i/-2i, 2/7-8i, 3-i/2-i, and... (answered by Fombitz)
Simplify each expression. 1.sqrt -81 2.(5-2i)(2+5i)... (answered by Gogonati,robertb)
Please help! I have so simplify each expression: 1. (2 + 3i) + (4+ 5i) 2. (5 +... (answered by solver91311)
I'm on algebra 2 and need help with my work please. It says find the value of the complex (answered by richwmiller)
2-i / 3-4i ,2-i/3-4i*3+4i/3+4i, 6+8i_3i_4i^2/15 ,10+5i/15 ,5(2+i)/5(3), 2/3 + 1/3i (answered by nerdybill,himakamel)