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# FP1 complex numbers question - need help! Watch

1. Hi guys! It's part (b) I am stuck on!

I have rationalised the denominator and got to

5a+(6-a^2)i all divided by 4+a^2

and from there I am completely stuck as to where to go...

Thank you!
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2. (Original post by iMacJack)
Hi guys! It's part (b) I am stuck on!

I have rationalised the denominator and got to

5a+(6-a^2)i all divided by 4+a^2

and from there I am completely stuck as to where to go...

Thank you!
just means that the imaginary part of z is the same as the real part of z.

i.e: Im(z) = Re(z). So equate those two parts and find a.

The reason why this is such is because the half-line making an angle of pi/4 with the x-axis is precisely the line y=x. Which is saying im(z) = re(z).
3. (Original post by Zacken)
just means that the imaginary part of z is the same as the real part of z.

i.e: Im(z) = Re(z). So equate those two parts and find a.

The reason why this is such is because the half-line making an angle of pi/4 with the x-axis is precisely the line y=x. Which is saying im(z) = re(z).
How does one think of that though? How does this help me find out the values of A?

Thanks again
4. (Original post by iMacJack)
How does one think of that though? How does this help me find out the values of A?

Thanks again
What is the imaginary part of your number in terms of a?

What is the real part of your number in terms of a?

Equate those to get an equation in only a and then solve for a.
5. (Original post by iMacJack)
How does one think of that though?
Try drawing a complex number on an argand diagram with .
6. Considering both the imaginary and real parts share the same denominator I can just equate the numerator bits can't I?

If so....

5a = 6-a^2

a^2+5a-6=0
(a+6)(a-1)=0
a=1 or a = -6

Okay - so now what? Do you think I should put those values in and see which one gives me the value in the first quadrant?

notnek Zacken
7. (Original post by iMacJack)
Considering both the imaginary and real parts share the same denominator I can just equate the numerator bits can't I?

If so....

5a = 6-a^2

a^2+5a-6=0
(a+6)(a-1)=0
a=1 or a = -6

Okay - so now what? Do you think I should put those values in and see which one gives me the value in the first quadrant?

notnek Zacken
You should be able to say a=1 right away since a=-6 will make 5a negative and not be in the first quadrant.
8. Cheers

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