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avatar+597 

Hi good people,

 

I did try this sum, but it becomes terribly horrible at some stage, and I just gave up...it says to calculate x and y if:

 

\(x+yi={3+2i \over 4-3i}\)

 

I rasionalised the denominator and carried on, but like I said, the sum just gets to much, and I fell out of the bus...please help!!..thanx again..

 May 20, 2019
 #1
avatar+6045 
+3

I hope you weren't too badly injured.

 

\(\dfrac{3+2i}{4-3i} = \\~\\ \dfrac{3+2i}{4-3i}\cdot \dfrac{4+3i}{4+3i} \\~\\ \dfrac{(12-6)+i(8+9)}{25}= \dfrac{6+17i}{25}\\~\\ \dfrac{6}{25}+\dfrac{17}{25}i\)

.
 May 20, 2019
 #2
avatar+597 
+2

haha, Hi Rom,

 

thanx for the sum, uhm, I do understand the part you did, what I fail to understand is that they ask to solve x and then y?

juriemagic  May 20, 2019
 #3
avatar+105705 
+1

They just want you to say 

 

x=6/25   and   y=17/25

Melody  May 20, 2019
 #4
avatar+597 
+1

huh???...golly me!!!...that's it???....Melody, ..thank you...hahahaha...yep, i certainly have a looong way to go...laugh

juriemagic  May 20, 2019
 #5
avatar+597 
+1

OOOHHHH!!!!!!... now I see it!!!!!!!!!!!...never!!!!!...I always tend to look too deep into things.....man!!!!

juriemagic  May 20, 2019
 #6
avatar+105705 
0

LOL      surprise

Melody  May 20, 2019

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