# Edexcel A Level Further Maths Core Pure 1 Complex numbers help

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#1
so i have the following question
(3√2+i) -(√2-i)
i am fine with the surd part of the question the problem is with the (i) and (-i) how do you deal with that for example if it was
(3√2+6i) -(√2-3i)
you would lay it out
(3√2-√2)- (6-3)i
which would obviously give you in the form a+bi where a,b are ∈ of ℝ
uou would get (3√2-√2)= 2√2 and (6-3)i you would multiply to get 3i

but when you have a question like (3√2+i) -(√2-i) when showing working would you just write out the imaginary part to the question as (+1-1)i or would you write it a different way i know how to do it just when it is a singular positive i or negative i in the question i don't know how to approach that
Last edited by liamlarner; 4 weeks ago
0
4 weeks ago
#2
(Original post by liamlarner)
so i have the following question
(3√2+i) -(√2-i)
i am fine with the surd part of the question the problem is with the (i) and (-i) how do you deal with that for example if it was
(3√2+6i) -(√2-3i)
you would lay it out
(3√2-√2)- (6-3)i
which would obviously give you in the form a+bi where a,b are ∈ of ℝ
uou would get (3√2-√2)= 2√2 and (6-3)i you would multiply to get 3i

but when you have a question like (3√2+i) -(√2-i) when showing working would you just write out the imaginary part to the question as (+1-1)i or would you write it a different way i know how to do it just when it is a singular positive i or negative i in the question i don't know how to approach that
(3√2-√2)- (6-3)i should be (3√2-√2) - [6-(-3)]i so 2√2 + 9i ??

for first question the imaginary part is [1-(-1)]i so 2i
0
#3
(Original post by golgiapparatus31)
(3√2-√2)- (6-3)i should be (3√2-√2) - [6-(-3)]i so 2√2 + 9i ??

for first question the imaginary part is [1-(-1)]i so 2i
Yeah what i thought representing them as 1 and minus 1 because it is i
0
4 weeks ago
#4
(Original post by liamlarner)
Yeah what i thought representing them as 1 and minus 1 because it is i
Sorry, not really sure what do you mean
If you have
z = a + bi
and
w = c + di

then
z + w = (a + bi) + (c + di) = (a+c) + (b+d)i

z - w = (a + bi) - (c + di) = (a-c) + (b-d)i
0
#5
(Original post by golgiapparatus31)
Sorry, not really sure what do you mean
If you have
z = a + bi
and
w = c + di

then
z + w = (a + bi) + (c + di) = (a+c) + (b+d)i

z - w = (a + bi) - (c + di) = (a-c) + (b-d)i
don't worry just saying i understand what you said in response to my original question my teacher tends to just use i and z don't know where w is from
1
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