SOLUTION: Imagnary numbers: {{{(1-3i)/(2+i)}}}

Algebra.Com
Question 20910: Imagnary numbers:

Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!
(1-3i)/(2+i)=...we rationalise the denominator by multiplying with its conjugate...conjugate of 2+i is 2-i...hence multiplying n.r and d.r with 2-i,
(1-3i)(2-i)/(2+i)(2-i)=[1*2-1*i-3i*2+(-3i)(-i)]/[2^2-i^2]=[2-7i-3]/(4+1)
=(-1-7i)/5=-(1+7i)/5....using (a+b)(a-b)=a^2-b^2 and i^2=-1

RELATED QUESTIONS

who descover imagnary numbers (answered by longjonsilver)
(2+3i)(1-i)= (answered by stanbon)
2-3i/1+i (answered by KMST)
Simplify (1-3i) + (2-i) + (2+i) +... (answered by ikleyn)
Simplify:... (answered by stanbon)
(3+3i/1-i)^2 (answered by stanbon)
The Sum of (3-the square root of 5) + (2-the square root of 45). We just where introduced (answered by ankor@dixie-net.com)
(2+3i)/(-4i)+(2+4i)/(1+i) (answered by CharlesG2)
solve: 2+i divided by... (answered by Fombitz)