What is the standard form of (2+3i)(4-7i)/(1-I)
2 answers:
Standard form is a+bi where a is the real part and bi is the imaginary part
we need to rationalize the denomenator
remember difference of 2 perfect squares
(a-b)(a+b)=a²-b²
so
in denom we got (1-i)
mulitply whole fraction by (1+i)/(1+i)
(2+3i)(4-7i)(1+i)/(1²-i²)
(2+3i)(4-7i)(1+i)/(1-(-1))
(2+3i)(4-7i)(1+i)/(1+1)
(2+3i)(4-7i)(1+i)/(2)
simplfy top part
(31+27i)/2
31/2+27i/2
the standard form is
we need to rationalize the denomenator
remember difference of 2 perfect squares
(a-b)(a+b)=a²-b²
so
in denom we got (1-i)
mulitply whole fraction by (1+i)/(1+i)
(2+3i)(4-7i)(1+i)/(1²-i²)
(2+3i)(4-7i)(1+i)/(1-(-1))
(2+3i)(4-7i)(1+i)/(1+1)
(2+3i)(4-7i)(1+i)/(2)
simplfy top part
(31+27i)/2
31/2+27i/2
the standard form is
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