Question

simplify: answer has to be arithmetic, simplified, no "i" in denominator, and in simplest form.

1) \[\frac{ 1 }{ (3i-1)(\frac{ 7 }{ 2 } i +3) }\]

2) \[\frac{ 7 }{ 3i+6 } \times \frac{ 4 }{ 9i(16+i ) }\]

Answer 1

@agent0smith @UsukiDoll

Answer 2

You can use conjugate to get the answer.
1)
\[\frac{ (3i+1)(\frac{ 7 }{ 2 }i-3) }{ (3i-1)(\frac{ 7 }{ 2 }i+3)(3i+1)(\frac{ 7 }{ 2 }i-3) }\]
\[= \frac{ -\frac{ 21 }{ 2 } - 3 - 5.5i }{ (9i^2 -1)(\frac{ 49 }{ 4 }i^2-9) }\]
\[= \frac{ -13.5 - 5.5i }{ (-10)(-21.25) }\]
\[= \frac{ -13.5 - 5.5i }{ 212.5}\]
\[= \frac{ -27 - 11i }{ 425}\]

You can do the same way with the second equation

Answer 3

O.O how did you get all that?

Answer 4

help...

Answer 5

know what conjugate is ?

Answer 6

no

Answer 7

conjugate of \(a-bi = a+bi \)
to find conjugate, you just flip the sign of imaginary term

Answer 8

uh...

Answer 9

and if you have a fraction like,
1/(a+bi)
in order to make the denominator as real, we multiply and divide by the conjugate of denominator because \((a+bi)(a-bi)= a^2 +b^2\)

Answer 10

you didn't get what conjugate is ?

Answer 11

ok right there i got lost

Answer 12

no i get that

Answer 13

just not this last one

Answer 14

know what 'i' is ?

Answer 15

imaginary

Answer 16

yes, i is an imaginary unit,
\(\large i = \sqrt {-1}\)
so can you tell me whats i^2 = ... ?

Answer 17

\[\sqrt{-1}^{2}\]

Answer 18

which equals ?

Answer 19

my calc says nan

Answer 20

lol
lets not use calculator :)
NAN means Not a Number (what your calc meant was, its not a REAL number, and we already know i is an imaginary number)

Answer 21

ok so....

Answer 22

hint :
\(\large (\sqrt a)^2 = a \\ \large (\sqrt {-1})^2 =...? \)

Answer 23

-1

Answer 24

ok

Answer 25

did some correction.

YES, i^2 = -1
remember this.

now lets go to your question,
before doing 1)
lets just try 1/(3i-1)

note that i multiplied numerator and denominator by conjugate of 3i-1
which is -3i+1