Complex Question Help ahhhhh!! (1 Viewer)

[DraconisV]

Ok you guys probably think this is easy, but i don't.
ok the first one is

(3 + 4i)(a + ib) = -2i, you have to find out what a and b are.

Ok, my answers are a= (4/3)b and b= -(4a+2)/3 I think these are wrong, if they are can someone please tell me the right answer and how to get it step by step.

The 2nd and last question i need help with (so far) is
(a + ib)^2 = 2i and you also have to find out what a and b are

My answers for this one are a=b and b=1/a i do feel this one is wrong for sure, can someone please show me how to do it please show me the answer and how you got it step by step.

thank you in advance

 

[haboozin]

Ok you guys probably think this is easy, but i don't.
ok the first one is

(3 + 4i)(a + ib) = -2i, you have to find out what a and b are.

Ok, my answers are a= (4/3)b and b= -(4a+2)/3 I think these are wrong, if they are can someone please tell me the right answer and how to get it step by step.

The 2nd and last question i need help with (so far) is
(a + ib)^2 = 2i and you also have to find out what a and b are

My answers for this one are a=b and b=1/a i do feel this one is wrong for sure, can someone please show me how to do it please show me the answer and how you got it step by step.

thank you in advance

(3 + 4i)(a + ib) = -2i, you have to find out what a and b are.

expand

3a - 4b + i(4a + 3b) = 0 -2i

so

equating reals and imaginary

3a - 4b = 0

4a + 3b = -2


SOLVE SIMLUTANIOUSLY


(a + ib)^2 = 2i and you also have to find out what a and b are

a^2 - b^2 + 2abi = 2i

so a^2 - b^2 = 0

2ab = 2


SOLVE SIMLUTANIOUSLY

 

[DraconisV]

Thx haboozin for the quick reply, hmmm now i must solve them simultaneously nice easy. thx

 

[Riviet]

Ok you guys probably think this is easy, but i don't.
ok the first one is

(3 + 4i)(a + ib) = -2i, you have to find out what a and b are.

Ok, my answers are a= (4/3)b and b= -(4a+2)/3 I think these are wrong, if they are can someone please tell me the right answer and how to get it step by step.

The 2nd and last question i need help with (so far) is
(a + ib)^2 = 2i and you also have to find out what a and b are

My answers for this one are a=b and b=1/a i do feel this one is wrong for sure, can someone please show me how to do it please show me the answer and how you got it step by step.

thank you in advance

The first one is not hard, here's how i did it:
(3+4i)(a+ib) = -2i
a+ib = -2i/(3 + 4i)
a+ib = -2i(3-4i)/(9+16) (multiply top and bottom by conjugate to cancel i in denominator)
a+ib = (-6i - 8)/25
= -6/25 i - 8/25
.: a = -8/25, b= -6/25

 

[Sepulchres]

OR (for the first one)

you could take (3 + 4i) to the other side, multiply un and down by its conjugate (ie. 3-4i) and then equate the real et imaginary parts. Either way works.

EDIT: Heh ^^ beat me to it.

 

[Riviet]

(3 + 4i)(a + ib) = -2i, you have to find out what a and b are.

expand

3a - 4b + i(4a + 3b) = 0 -2i

so

equating reals and imaginary

3a - 4b = 0

4a + 3b = -2


SOLVE SIMLUTANIOUSLY


(a + ib)^2 = 2i and you also have to find out what a and b are

a^2 - b^2 + 2abi = 2i

so a^2 - b^2 = 0

2ab = 2


SOLVE SIMLUTANIOUSLY

Wow, thats an interesting way to do it. *claps*

OMG!! I just looked at the second question and got a feel of deja vu! I swear i was doing that question last night for hw. What textbook are you using Draconis?

 

[DraconisV]

What textbook are you using Draconis?

Well, i got this question from my teacher in a homework sheet, so he just makes em up. BTW im using some fitzpatrick book which has like a brownish-beige kinda cover and it says that it was printed in 1992 and has been used since then, should i be using a more up to date one.

 

[haboozin]

Wow, thats an interesting way to do it. *claps*

isnt this how everyone does it...

kinda logical..

i think it is
its how hsc solutions do it..

 

[DraconisV]

well, i did the simultanious one with the first one and yay i got the first one right now.

but im having trouble with the second one

with the a^2 - b^2 = 0 and 2ab=2 (mmn, wouldn't the second one be the same as ab=1 as you just divide it by 2)

So how you do it, mmmn confused again.

Also are these really easy complex number questions (of this format), like for example could you get Find a and b where a + ib = 2cosec(theta)^2 + 6(theta)i where theta is 45 degrees lol, this question doesnt need an answer, so would you be able to get somethign that insane.

 

[Stefano]

well, i did the simultanious one with the first one and yay i got the first one right now.

but im having trouble with the second one

with the a^2 - b^2 = 0 and 2ab=2 (mmn, wouldn't the second one be the same as ab=1 as you just divide it by 2)

So how you do it, mmmn confused again.

Also are these really easy complex number questions (of this format), like for example could you get Find a and b where a + ib = 2cosec(theta)^2 + 6(theta)i where theta is 45 degrees lol, this question doesnt need an answer, so would you be able to get somethign that insane.

1. 2ab=2 ===> ab=1

2. Yes, these are relatively easy 'complex' number questions.

 

[Antwan23q]

ohh i see, this is for someone who just started... ok i was gettin worried that this was someone with the hsc next week...

 

[Riviet]

ohh i see, this is for someone who just started... ok i was gettin worried that this was someone with the hsc next week...

Lol! They would be screwed for the exam

 

[DraconisV]

yes i have only just started the hsc 4 unit course (like i was in yr 11 few weeks ago), if i was doing my hsc in a few weeks, I would just kill myself if i didnt know how to do them.

 

[yrtherenonames]

Don't kill yourself man. You've got an HSC to do

 

[Stefano]

Be like the bumblebee DraconisV!

(That rhymes)

 

[DraconisV]

Be like the bumblebee DraconisV!

(That rhymes)

Funny you say that Stefano, coz the bumblebee was used in my previous sig. hmm where did that sig go, bumble bumble where are you, hmm