# complex numbers

#1
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
Last edited by dude_idk; 1 month ago
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1 month ago
#2
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
Last edited by NiceMan420; 1 month ago
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1 month ago
#3
(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?
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#4
(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
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#5
(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
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1 month ago
#6
(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?
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#7
(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
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1 month ago
#8
(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.
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#9
(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?
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1 month ago
#10
(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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