Question

write the expression as a complex number in standard form
1+I/2-3i

Answer 1

\[\frac{1+i}{2-3i}\]To write this in standard form, start by rationalizing it. Multiply the numerator and denominator by the conjugate of the denominator, which is \(2+3i\). After doing so, the denominator will be a difference of squares, with the imaginary term becoming a real term after substituting \(i^2 = -1\).

Answer 2

so it will be 2+3i over 1i

Answer 3

Well, ignoring any questions about whether you did the algebra correctly, is that a complex number in standard form?

Answer 4

\[(2-3i)(2+3i) =\]

Answer 5

i dont understand

Answer 6

Can you do that multiplication?

Answer 7

\[(2-3i)(2+3i) = 2(2+3i)-3i(2+3i) = 2*2 + 2*3i -2*3i-3i*3i\]\[=4+6i-6i-9i^2 = 4-9i^2\]
What do you get when you substitute \(i^2=-1\) into that result?

Answer 8

you get -9

Answer 9

No. 4-(-9) = ?

Answer 10

13

Answer 11

Right. so the new denominator is 13. The new numerator is \[(1+i)(2+3i)\]What does that equal?

Answer 12

3+4i

Answer 13

No. How did you arrive at that?

Answer 14

You have to multiply, not add

Answer 15

This is just like multiplying (a+b)(c+d).

\[(1+i)(2+3i) = 1(2+3i) + i(2+3i) =\]

Answer 16

do you multiply like terms together

Answer 17

can you mutiply 3i and 1

Answer 18

Yes, of course. How do you think you could form \(3i\) if you couldn't multiply a real with an imaginary? :-)

Answer 19

Just pretend \(i\) is like \(x\) (or any other letter) for the purposes of doing the multiplication.

Answer 20

can you get a x to the 2nd power in this equation because we're using x

Answer 21

\[(1+i)(2+3i) = 1(2+3i) + i(2+3i) = 1*2 + 1*3i + 2*i + i*3i\]Can you simplify that?

Answer 22

im confused

Answer 23

Do you understand how to multiply binomials?

Answer 24

yeah when there like terms

Answer 25

\[(a+b)(c+d)\]? That's exactly what we are doing here, it's just that one of our letters has a special meaning.

Answer 26

Here. Can you multiply
\(1+a)(2+3a)\]for me?

Answer 27

2+4a

Answer 28

No. Show me your work...

Answer 29

you multiply 1 and 2 cause their the same then a and 3a

Answer 30

No. Can you multiply 12*15?

12
15
----
60
120
-----
180

That's exactly the same thing as multiplying
(5+10)(2+10) = 5*2 + 5*10 + 10*2 + 10*10 = 10+50+20+100 = 180

By your system, 12*15 = (2+10)*(5+10) = 2*5 + 10*10 = 110.

Answer 31

Distributive property of multiplication:

a(c+d) = a*c + a*d = ac + ad

(a+b)(c+d) = a(c+d) + b(c+d) = a*c + a*d + b*c + b*d = ac+ad+bc+bd

Answer 32

where did you get 60 and 120

Answer 33

Each piece of each multiplier has to be multiplied by each piece of the the other.

\[(1+i)(2+3i) = 1*(2+3i) + i*(2+3i)\] (distributive property)
\[1*(2+3i)+i*(2+3i) = 1*2 + 1*3i + 2*i + i*3i \](distributive property, again)
\[1*2+1*3i+2*i + i*3i = 2 + 3i+2i+3i^2 = 2+5i+3i^2\](collecting like terms)
\[2+ 5i+3i^2 = 2+5i+3(-1) = 2-3+5i=-1+5i\](using definition of \(i^2=-1\))

Answer 34

5*12 = 60
10*12 = 120
How would you have done 12*15 without a calculator?

Answer 35

i did it with a calculator you said 12*15=180

Answer 36

Don't you know how to do it without a calculator?

Answer 37

thats how i know how to do it

Answer 38

Your education has done you a disservice, I'm afraid. If you don't understand how to multiply polynomials, or even do multiplication by hand, it will be quite a slog :-(

Well, I can't fix that. I can show you the entire problem "we" did, however:

\[\frac{(1+i)}{(2-3i)} = \frac{(1+i)(2+3i)}{(2-3i)(2+3i)} \]multiply numerator and denominator by conjugate of denominator
\[\frac{(1+i)(2+3i)}{(2-3i)(2+3i)} = \frac{1*2+1*3i+2*i+i*3i}{2*2+2*3i-3i*2-3i*3i} = \frac{2+5i+3i^2}{4-9i^2}\] Now we substitute \(i^2=-1\)
\[\frac{2+5i+3(-1)}{4-9(-1)} = \frac{-1+5i}{13} = -\frac{1}{13}+\frac{5i}{13}\]And that's our expression in standard form.

Answer 39

so your saying im not smart

Answer 40

No, I'm saying you don't know a bunch of very important stuff, and it will be a real challenge to do well.

Answer 41

Are you studying math in an online course, or do you have a teacher you can talk to?

You would probably benefit from watching the Khan Academy videos on multiplying monomials, binomials, and polynomials. The first one is here: https://www.khanacademy.org/math/algebra/polynomials/multiplying_polynomials/v/multiplying-monomials

Answer 42

school is out for the summer so no ones at my school

Answer 43

Okay, that's unfortunate, but on the flip side, you don't have a lot of homework to fill your time :-) I would give serious consideration to spending an hour each day going through this material (watching the videos, borrowing a book from the local library, etc.). None of this is particularly difficult, but it does require understanding what you are doing, and attention to detail.

Answer 44

I would encourage you to swallow your pride and borrow a 4th or 5th grade math book from the children's section of the local library, and learn how to do multiplication and division without a calculator.

Answer 45

ok what websites your i use

Answer 46

https://www.khanacademy.org/math/algebra would be an excellent start.