Complete question:
What values of c and d would make the following expression represent a real number? i(2 + 3i)(c + di)
choices: A)c = 2, d = 3 B)c = –2, d = 3 C)c = 3, d = –2 D)c = –3, d = –2
Option D is correct. The value of c and d for the expression to be real are -3 and -2 respectively
Given the expression
[tex]i(2 + 3i)(c + di)[/tex]
Expand the expression
[tex]=i(2 + 3i)(c + di)\\=(2i+3i^2)(c+di)\\=(2i+3(-1))(c+di)\\=(2i-3)(c+di)\\=2ic+2di^2-3c-3di\\=2ic-2d-3c-3di[/tex]
Collect the like terms
[tex]=2ic-3di-2d-3c[/tex]
For the resulting function to be real, then the imaginary part must be zero i.e.
[tex]2ic-3di=0\\2ci = 3di\\2c = 3d\\c=\frac{3d}{2}\\[/tex]
If d = -2, then;
[tex]c=\frac{3(-2)}{2}\\c=\frac{-6}{2}\\c= -3[/tex]
Hence the value of c and d for the expression to be real are -3 and -2 respectively.
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