Question

What is the quotient of the expression given in standard form?

Answer 1

\[\frac{ 1+2\sqrt{3i} }{ 1+\sqrt{3i} }\]

Answer 2

rationalize by multiplying times the denominator's conjugate.

Answer 3

what is the conjugate for \(1+\sqrt{3i}\) ?

Answer 4

1-sqrt(3u)

Answer 5

(3i)

Answer 6

that u was supposed to by "i" correct?

Answer 7

yes

Answer 8

Yes, that is the correct conjugate

Answer 9

yea

Answer 10

ok, now multiply

Answer 11

wait, is it \(\large\color{black}{ \displaystyle \frac{1+2\sqrt{3i}}{1-\sqrt{3i}} }\)

or is it \(\large\color{black}{ \displaystyle \frac{1+2i\sqrt{3}}{1-i\sqrt{3}} }\)

Answer 12

\[\frac{ 7 }{ 4 }-\frac{ \sqrt{3} }{ 4}i\]

Answer 13

you did this for the first or the second expression that I offered?

Answer 14

I think this was the first one

Answer 15

in the first one it is not an i, it is the 4th root of -1 , not a second.

Answer 16

\[\frac{ 7 }{ 4 }+\frac{ \sqrt{3} }{ 4 }i\]

Answer 17

I promise I will fall asleep if I stop now

Answer 18

\(\large\color{black}{ \displaystyle \frac{1+2\sqrt{3i}}{1+\sqrt{3i}} }\)

(yeah, a +, not a - )

\(\large\color{black}{ \displaystyle \frac{(1+2\sqrt{3i})(1-\sqrt{3i})}{(1+\sqrt{3i})(1-\sqrt{3i})} }\)

\(\large\color{black}{ \displaystyle \frac{1+\sqrt{3i}-6i}{1-3i} }\)

\(\large\color{black}{ \displaystyle \frac{(1+\sqrt{3i}-6i)(1+3i)}{(1-3i)(1+3i)} }\)

I will post till this point now

Answer 19

\(\large\color{black}{ \displaystyle \frac{1+\sqrt{3i}-6i+3i+3i\sqrt{3i}+18}{1+3} }\)

Answer 20

did I post the answer?

Answer 21

\(\large\color{black}{ \displaystyle \frac{19+(1+3i)\sqrt{3i}-3i}{4} }\)

Answer 22

I didn't even get the question

Answer 23

this is the first expression, and it doesn't really have a simple \(a+bi\) form

Answer 24

\[\frac{ 7 }{ 4 }+\frac{ \sqrt{3} }{ 4 }i \]

Answer 25

what problem were you doing, were you doing \(\large\color{black}{ \displaystyle \frac{1+2\sqrt{3i}}{1+\sqrt{3i}} }\) ?

Answer 26

then, it is incorrect, b/c there is not standard form for this expression.

Answer 27

On the other hand, if it is \(\large\color{black}{ \displaystyle \frac{1+2i\sqrt{3}}{1+i\sqrt{3}} }\) ....

Answer 28

\(\large\color{black}{ \displaystyle \frac{(1+2i\sqrt{3})(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})} }\)

\(\large\color{black}{ \displaystyle \frac{1+i\sqrt{3}-2(-1)(3)}{1-(-1)(3)} }\)

\(\large\color{black}{ \displaystyle \frac{1+i\sqrt{3}+6}{1+3} }\)

\(\large\color{black}{ \displaystyle \frac{7+i\sqrt{3}}{4} }\)

Answer 29

then you are correct....

Answer 30

:)