please someone tell me how to do this, I need to know step by step?

Question

9+7i/ 9+3i
divide I dont get this imanginary numbers and I need to know step by step.. please..

Answer 1

(9+7i)(9-3i)/(9+3i)(9-3i)
=(81+21+36i)/(81+9)
=102/90+(36/90)i
=(51/45)+(2/5)i

Answer 2

You have to get imaginary numbers out of the denominator (Like you do with radicals) so you are going to have to multiply by 1. Take your equation and multiply by the negative like this
(9+7i) x (9-3i) (As you can see, 9-3i / 9-3i = 1)
(9+3i) (9-3i)
Multiplying your two denominators will end up cancelling out the i.
You multiply 9 by 9 adn then 3i by -3i. 9x9=81, 3ix-3i = 9i /\2(squared). i squared gives you -1 (there are rules for this). So now, 81 - (-9) = 90. That is your final denominator.

As for the numerator, you have to distribute using FOIL (first outer inner last).
So, 9 x 9=81, 9 x -3i = -27i, 7i x 9 =63i, and 7i x -3i = -21 i /\2 . The -21 i squared then = +21 (because -21 x -1).

Okay so your numerator is 102 + 36 i
and your denominator is 90
but you have to reduce (all numbers can be divided by 6)

You answer should be: (17 + 6 i ) / 15.

Check the math (i did the multiplication and division in my head pretty quickly) but the steps are correct - I've taught this subject!

Answer 3

Dividing by an imaginary number is a bit trickier than dividing by a real one.

To divide by an imaginary number, you have to make the denominator a real number. Which means you have to multiply by it's complex conjugate. In order to do that you need to know about complex conjugates.

Two imaginary numbers are complex conjugates if their sum is a real number, and their product is a nonnegative real number. In other words for imaginary number a + bi, it's complex conjugate is a - bi.


So back to your problem:

(9+7i)/ (9+3i)

First multiply the above by (9+3i)/ (9+3i). (You have to multiply by a form of 1. You don't want to change the value of the expression now do you?)

You multiply the imaginary numbers like polynomials, with FOIL (first, outer, inner, last) and the only tricky part is you have to remember about are the i's (i^2 = -1). After doing this you get:

((9+7i)(9-3i))/ ((9+3i)(9-3i))

Simplifying the expression with FOIL, you get:

(81 + 63i - 27i - 21(i^2))/ (81 + 27i -27i - 9(i^2)) -> Simplify further:

(81 + 36i + 21)/ (81 + 9) -> Simplify Once More...

(102 + 36i)/ (90) -> And finally rearrange it to imaginary form:

102/90+ (36/90)i -> More Simplifying...

17/15 + (2/5)i -> Ta Da!

Answer 4

Okay, what you do is multiply the series by the denominator over itself, only make the positive a minus (the opposite of the imaginary number); so you multiply the original by (9-3i)/(9-3i). When you multiply, you FOIL (I don't if that's universal term or not...) and get (81+36i-21i^2)/(81-9i^2).

Now, understand that i=the square root of -1, so i^2 equals the square root of -1 squared. Sooooo, the square root and square exponent cancel each other, so (i^2)=(-1).

So now you can simplify the product of the first multiplication and get (80+36i)/90.

Hope that wasn't too complicated... It's kinda hard to explain without writing it out. Let me know if you still don't get it and maybe I can draw it in paint or something and email it to you.

Answer 5

The trick is to get rid of the denominator (9+3i). You do this by multiplying and dividing by the conjugate of the denominator, which is 9-3i

9+7i (9+7i) (9-3i)
------ = ------ x -------
9+3i (9+3i) (9-3i)

you can do this because 9-3i/9-3i = 1 so it doesnt change your answer. now you have (9+7i)(9-3i) / (9+3i)(9-3i) which you should be able to solve easier.

Answer 6

9 + 7i/9 + 3i =

9 = 7i/ 9 + 3i x 9 - 3i / 9 - 3i =

81 + 63i - 27i - 21i²

The numerator equation becomes

81 + 63i + 21=

102 + 63i

Multiplying the denominator

(9 + 3i)(9 - 3i) =

81 + 27i - 27i - 9i²

The equation becomes

81 + 9 = 90

- - - - - - - - - - - - - -

102/90 + 63i/90

- - - - - -s-

Answer 7

(9+7i)/(9+3i)
To simplify this you need to multiply both numerator and denominator 9-3i. 9-3i is called the complex conjugate of 9+3i.

So we get [(9+7i)(9-3i)]/[(9+3i)(9-3i)]
=[81 +24i +63i +21i^2]/(81+9)
=[60 + 87i]/90 = 2/3 +( 29/30)i

You should note tha i^2= -1 and that (9+3i)(9-3i)= 9^2 + 3^2
In general (a+bi)(a-bi)= a^2 + b^2

Answer 8

Clearly, you can mulitply both numerator and denominator by any non-zero number without changing anything. Do so, and let the number be the complex conjugate of the denominator. This will turn the denominator into a real number, and the numerator will be complex but you can probably reduce the fractions.

Answer 9

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Answer 10

treat i like you would any other letter.
9+7i/9+3i=9+7i/9+27i/9=9+34i/9