Asked by Nayla
Does (-7/4i-1) symplify to be -7(1+4i)/18
Answers
Answered by
drwls
Not quite. Your parenthese seem to be in the wrong place, and your math is wrong.
-7/(4i-1)= -7(4i+1)/[(4i+1)(4i-1)]
= -7(4i +1)/(-17)
= (7/17)(1+4i)
-7/(4i-1)= -7(4i+1)/[(4i+1)(4i-1)]
= -7(4i +1)/(-17)
= (7/17)(1+4i)
Answered by
Nayla
These are complex numbers by the way.
Answered by
Nayla
Thanks
Answered by
Nayla
why is the 1+4i on the side of the fraction and not on the top with the 7
Answered by
MathMate
(7/17)(1+4i) is the same as 7(1+4i)/17.
I too think it is simpler to write it as DRWS did, where the numeric fraction precedes the algebraic part.
I too think it is simpler to write it as DRWS did, where the numeric fraction precedes the algebraic part.
Answered by
Jaison
is (7/17)-(28/17i) the same thing
Answered by
MathMate
"is (7/17)-(28/17i) the same thing "
No it is not.
(7/17)*(1+4i)
=(7/17)*1 + (7/17)*4i
=(7/17) + 28i/17
But this is not a simplified version.
In any case the object is to remove the "i" from denominator, or "rationalizing the denominator". Some teachers may consider it incomplete if square-root signs or "i" appear in the denominator.
The reason that the numerator and denominator were multiplied by (1+4i) was to create the product (1+4i)*(1-4i)=-17 in the denominator, thereby rationalizing it.
No it is not.
(7/17)*(1+4i)
=(7/17)*1 + (7/17)*4i
=(7/17) + 28i/17
But this is not a simplified version.
In any case the object is to remove the "i" from denominator, or "rationalizing the denominator". Some teachers may consider it incomplete if square-root signs or "i" appear in the denominator.
The reason that the numerator and denominator were multiplied by (1+4i) was to create the product (1+4i)*(1-4i)=-17 in the denominator, thereby rationalizing it.
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