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Select the correct answer.
What is the standard form of '((2+3i)(4-71))/((1-1)) ?
O A.
-3/13-4/261
OB.
*3/13+4/261
O c.
31/2+27/21
OD.
.-31/2-27/21
Plz halp

Sagot :

Answer:

[tex]\dfrac{31}{2}+\dfrac{27}{2}i[/tex]

Step-by-step explanation:

The given complex number is:

[tex]z=\dfrac{(2+3i)(4-7i)(1+i)}{(1^2-i^2)}[/tex]

Firstly we can solve (2+3i)(4-7i)(1+i)

= 2(8)+2(-7i)+4(3i)+3i(-7i²)× (1+i)

= (16-14i+12i-21i²i)× (1+i)

Since, i² = -1

= (16-14i+12i-21(-1)i)× (1+i)

= (16-14i+12i+21)× (1+i)

= (29-2i)(1+i)

= 31+27i

Denominator : (1-i²)

= (1-(-1))

= 2

[tex]\dfrac{(2+3i)(4-7i)(1+i)}{(1^2-i^2)}=\dfrac{31+27i}{2}\\\\=\dfrac{31}{2}+\dfrac{27}{2}i[/tex]

Hence, the correct answer is [tex]\dfrac{31}{2}+\dfrac{27}{2}i[/tex].

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