Question

Algebra 2 help, Simplify the expression;
sqrt(-1) / (2 -4i)-(5+3i)

Answer 1

Is this the problem?

\(\dfrac{\sqrt{-1}}{(2 - 4i) - (5 + 3i)} \)

Answer 2

I can attach a image or draw it out if needed

Answer 3

That means it's not what I asked?

Answer 4

Can you show what the problem is supposed to be if I didn't understand it correctly?

Answer 5

\[\frac{ \sqrt{-1} }{ (2-4i)-(5+3i) }\]

Answer 6

Ok. Start by simplifying \( \sqrt{-1} \)

Answer 7

i

Answer 8

Good.

Answer 9

Now you need to do the subtraction of complex numbers in the denominator.
Drop the first set of parentheses because it is unnecessary.
Then distribute the negative to the left of the second set of parentheses.

Answer 10

2-4i - (-)5-3i

Answer 11

-3-7i

Answer 12

\(\dfrac{ \sqrt{-1} }{ (2-4i)-(5+3i) }\)

\(= \dfrac{ i}{ 2-4i-(5+3i) }\)

Be careful with the distribution of the negative:

\(= \dfrac{ i}{ 2-4i-5-3i }\)

Answer 13

The 5 inside the parentheses is positive, so it becomes negative, or -5.
The 3i is positive, so it becomes -3i.

Answer 14

So would it be -7-7i? or -3-7i?

Answer 15

Now you combine like terms in the denominator.

Answer 16

\(= \dfrac{ i}{ -3-7i }\)

Answer 17

So now you use the conjugate to eliminate the i, right

Answer 18

Now we need one more step, because we're not supposed to leave i in the denominator.

Answer 19

Correct, but it's to eliminate i from the denominator. There can be i in the numerator.

Answer 20

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

Answer 21

Be careful with the complex conjugate. The only sign that changes is the sign on the term with i.

\(= \dfrac{ i}{ -3 - 7i } \times \dfrac{-3 + 7i}{-3 + 7i}\)

Answer 22

Oh wow, I think that's where i've been going wrong

Answer 23

The conjugate of a + bi is a - bi, not -a - bi.
In fact what you did, by changing both signs, you are using the opposite (also called the additive inverse) of the complex number. That is not the complex conjugate.

Answer 24

Now you can multiply out the numerators and denominators.

Answer 25

\[\frac{ -3i-7 }{ 58 }\]

Answer 26

\(= \dfrac{ -3i + 7i^2 }{(-3)^2 - (7i)^2}\)

\(= \dfrac{ -3i - 7 }{9 - 49(-1)}\)

\(= \dfrac{-7 - 3i}{9 + 49} \)

\(=\dfrac{-7 - 3i}{58} \)

\(=-\dfrac{7}{58} - \dfrac{3}{58}i\)

Answer 27

You are correct.

Answer 28

Thank you so much!