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FP1 complex numbers question - need help!

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!
piby4.png15.3KB
Reply 1
Avatar for Zacken
Zacken
22
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!


arg⁡z=π4\arg{z} = \frac{\pi}{4} 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).
Reply 2
Avatar for iMacJack
iMacJack
OP
16
Original post by Zacken
arg⁡z=π4\arg{z} = \frac{\pi}{4} 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 :smile:
Reply 3
Avatar for Zacken
Zacken
22
Original post by iMacJack
How does one think of that though? How does this help me find out the values of A?

Thanks again :smile:


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.
Reply 4
Avatar for Notnek
Notnek
21
Original post by iMacJack
How does one think of that though?

Try drawing a complex number zz on an argand diagram with arg⁡z=π4\displaystyle \arg{z} = \frac{\pi}{4}.
Reply 5
Avatar for iMacJack
iMacJack
OP
16
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
Reply 6
Avatar for Zacken
Zacken
22
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.
Reply 7
Avatar for iMacJack
iMacJack
OP
16
Cheers

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