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# Further pure maths imaginary numbers watch

1. I can do parts a and b but c has stumped me. Can anyone help?
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2. (Original post by CPW2016)
I can do parts a and b but c has stumped me. Can anyone help?
What have you tried? - please post your working. Notice that the left hand side of c) are the two fractions you've simplified in b) multiplied by A and B respectively.
3. (Original post by notnek)
What have you tried? - please post your working. Notice that the left hand side of c) are the two fractions you've simplified in b) multiplied by A and B respectively.
i did notice the two fractions. I main query is: part c contains one equation but we have two unknowns. i dont know how i would find A and B without knowing at least one other equation involving z1 and z2
4. (Original post by CPW2016)
i did notice the two fractions. I main query is: part c contains one equation but we have two unknowns. i dont know how i would find A and B without knowing at least one other equation involving z1 and z2
It's one complex equation, which is enough to find two real unknowns.
5. (Original post by CPW2016)
i did notice the two fractions. I main query is: part c contains one equation but we have two unknowns. i dont know how i would find A and B without knowing at least one other equation involving z1 and z2
If you simplify then you should be able to compare real and imaginary parts on either side of the equation. This should generate two simulatenous equations for A and B (one for real parts and the other for imaginary parts).
6. (Original post by notnek)
If you simplify then you should be able to compare real and imaginary parts on either side of the equation. This should generate two simulatenous equations for A and B (one for real parts and the other for imaginary parts).
Thank you; i cant believe i didn't see this.

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