Complex Numbers Question

I'm having problems with this question. Can I get some help? Edexcel A level further Maths Complex Numbers.

I've converted into mod arg form and I get:

64 (cos(π/3) + isin(π/3)).

Upon taking the cube root, I get 4(cos(π/9) + sin(π/9)) through de Moivre's theorem, which is the same as 4e^(iπ/9).

I then add 2π/3 for each root, which then gives me 4e^(7π/9) and then 4e^(-5π/9), since the argument has to be between -π and π.

However, the mark scheme says this:

Screenshot 2024-04-26 at 19.34.53.jpg

.

I can't see what's wrong with my working so if you could please point it out, it would be much appreciated.

Edit: Sorry about the mark scheme. I can't make it any clearer and it's exactly how it appears for me. This is from the complex numbers year 2 topic questions on Physicsandmathstutor for Edexcel A level.

(edited 7 months ago)

@mqb2766 Any ideas?

Youve not said what your mod-arg number represents but if its z^3 then youve done the classic arctan mistake. Both the real and imaginary parts are negative (take the 32... over) and so the arg is -2pi/3 not pi/3. arctan is only valid for -pi/2..pi/2.

(edited 7 months ago)

Reply 3

vnayak

Original post by mqb2766

I'm not sure I fully understand. When I take the 32 and the 32i sqrt(3) over to the other side, I get z^3 = -32 - 32i sqrt(3).

the argument is tan ø = imaginary part/real part

tan ø = 32 sqrt(3) / 32

tan ø = sqrt (3) so ø = π/3, which is exactly what they get in the mark scheme (notes section for M1A1). Now, the root for ø they've used is just π/3 ± π so I am close but I'm not sure how you get -2π/3.

The arctan I've worked out also falls into the range that you've specified (-π/2 to π/2).

Reply 4

mqb2766

Original post by vnayak

I'm not sure I fully understand. When I take the 32 and the 32i sqrt(3) over to the other side, I get z^3 = -32 - 32i sqrt(3).
the argument is tan ø = imaginary part/real part
tan ø = 32 sqrt(3) / 32
tan ø = sqrt (3) so ø = π/3, which is exactly what they get in the mark scheme (notes section for M1A1). Now, the root for ø they've used is just π/3 ± π so I am close but I'm not sure how you get -2π/3.
The arctan I've worked out also falls into the range that you've specified (-π/2 to π/2).

Both the numbers 1+i and -1-i have the same arg by your arg(ument). A quick sketch in the argand plane should show theyre different by pi (+/-). The imag/real gives a gradient of the line, but youre after the arg of the halfline corresponding to the number. Its one of those basic things its easy to forget/overlook.

You do cs, so look up what the function arctan2() does and why you specify the x and y arguments seperately, not a single y/x.

(edited 7 months ago)

Reply 5

vnayak

Original post by mqb2766

Both the numbers 1+i and -1-i have the same arg by your arg(ument). A quick sketch in the argand plane should show theyre different by pi (+/-). The imag/real gives a gradient of the line, but youre after the argument of the halfline corresponding to the number. Its one of those basic things its easy to forget/overlook.
You do cs, so look up what the function arctan2() does and why you specify the x and y arguments seperately, not a single y/x.