Verify if (4+3i)/(3-4i)-(2+i)/(1-2i)

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hala718 eNotes educator| Certified Educator

let z= (4+3i)/(3-4i) - (2+i)/(1-2i)

First lest us simplify:

==> z= (4+3i)(3+4i)/(3-4i)(3+4i)  - (2+i)(1+2i)/(1-2i)(1+2i)

==> z= (12+25i-12)/(3^2 +4^2) - (2+3i -2)/(1^2 + 2^2)

==> z= 25i/25 - 3i/3

==> z= i -i = 0

==> z= 0 + 0i  

neela | Student

To verify (4+3i)/3-4i) = (2+i)/(1-2i) corrected.

In each case we realise the denominator by multiplying by the conjugate of the denominator. Conjugate of (x+yi)  is (x-yi).

LHS= (4+3i)/(3-4i) = (4+3i)(3+4i)/(3-4i)(3+4i)

= {12 +(16+9)i+12i^2}/ {9-16i^2)

={12-12+25i}/{9+16} as i^2 = -1

= 25i/25 = i.

RHS: (2+i)/(1-2i) = (2+i)(1+2i)/(1-2i)(1+2i)

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

Therefore LHS = RHS.

So the given equality verifies.

giorgiana1976 | Student

To verify if the expression is a complex number, we'll have to calculate it.

First, we'll calculate the least common denominator of the ratios.

LCD = (3-4i)(1-2i)

We'll multiply the 1st ratio by (1-2i) and the 2nd ratio by (3-4i) and we'll get:

 [(4+3i)(1-2i) - (2+i)(3-4i)]/(1-2i)(3-4i)

We'll remove the brackets and we'll have:


We'll eliminate and combine like terms form numerator:

0/(1-2i)(3-4i) = 0

We notice that the expression is a real number. We can also write it as a complex number, too.

0 = 0 + 0*i

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