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complex numbers

how does (3+2i)^2 equal 5+12i. like i can see that it may be the addition of 3+2 for the 5 and the 2(3x2) for the 12i but where does that come from and why doesnt it follow the normal (a+b)^2 = a^2+2ab+b^2
(edited 1 year ago)
I have no idea how much algabraic expansion you have done but to expand this equation you would:
(3+2i)^2 = (3+2i)(3+2i)
Expanding it gives 9 + 6i + 6i + 4(i^2) If you have never encountered this search "expanding double brackets"
Now we know that i = sqrt(-1) so naturally i^2 is just -1 so the equation becomes 9 + 6i + 6i +4(-1) -> 9 + 6i + 6i -4
Which, collecting like terms, makes 5 + 12i
(edited 1 year ago)
Reply 2
Original post by dude_idk
how does (3+2i)^2 equal 5+12i. like i can see that it may be the addition of 3+2 for the 5 and the 2(3x2) for the 12i but where does that come from and why doesnt it follow the normal (a+b)^2 = a^2+2ab+b^2

It does still follow the normal (a+b)^2. Do you know what i^2 is equal to?
Reply 3
Original post by Skiwi
It does still follow the normal (a+b)^2. Do you know what i^2 is equal to?

i do yeah but the answer sheet says 5+12i
Reply 4
Original post by NiceMan420
I have no idea how much algabraic expansion you have done but to expand this equation you would:
(3+2i)^2 = (3+2i)(3+2i)
Expanding it gives 9 + 6i + 6i + 4(i^2) If you have never encountered this search "expanding double brackets"
Now we know that i = sqrt(-1) so naturally i^2 is just -1 so the equation becomes 9 + 6i + 6i +4(-1) -> 9 + 6i + 6i -4
Which, collecting like terms, makes 5 + 12i

ohhhh perfect i think i just used a minus sign in the wrong place thank you
Reply 5
Original post by dude_idk
i do yeah but the answer sheet says 5+12i

What do you get when you expand (3+2i)^2 using a^2+2ab+b^2? Given you know i is the sqrt(-1) then what would i^2 be? Can you substitute anything into the expression you have?
Reply 6
Original post by dude_idk
i do yeah but the answer sheet says 5+12i


Original post by Skiwi
What do you get when you expand (3+2i)^2 using a^2+2ab+b^2? Given you know i is the sqrt(-1) then what would i^2 be? Can you substitute anything into the expression you have?

yeah so the og question was if 3+2i is a root of z^2+pz+q find p and q so i assumed since the other root is 3-2i you multiply out those two in which i just got 15
Reply 7
Original post by dude_idk
yeah so the og question was if 3+2i is a root of z^2+pz+q find p and q so i assumed since the other root is 3-2i you multiply out those two in which i just got 15

If i told you that and equation had roots x=5 and x=3 could you tell me what the original equation was by working backwards? Try applying the same process when the roots are z=3+2i and z=3-2i. You're correct in your assumption of the other root being 3-2i, the two complex solutions to a quadratic are conjugates, 3-2i/3+2i , 5+9i/5-9i , -2+6i/-2-6i etc.
Reply 8
Original post by Skiwi
If i told you that and equation had roots x=5 and x=3 could you tell me what the original equation was by working backwards? Try applying the same process when the roots are z=3+2i and z=3-2i. You're correct in your assumption of the other root being 3-2i, the two complex solutions to a quadratic are conjugates, 3-2i/3+2i , 5+9i/5-9i , -2+6i/-2-6i etc.

x^2-8x+15. so did i go wrong in my multiplication of 3-2i and 3+2i is it not 15?
Reply 9
Original post by dude_idk
x^2-8x+15. so did i go wrong in my multiplication of 3-2i and 3+2i is it not 15?

Think i may be misunderstanding what you're asking for. Are you just asking if (3-2i)(3+2i) is 15? or do you need help with the general question of what the original equation was?

It isn't 15 though, you've probably made a small slip up somewhere

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