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.

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:

(4-8i+3i+6-6+8i-3i-4)/(1-2i)(3-4i)

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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