# Engineering complex numbers + trig help

Got three questions to do, I’ve attempted them all and have an answer for 2 & 3 but no idea if they’re right. But I’m not sure how to continue question 1? I’ve found the polar form of 6+2j but not sure how to actually find the roots and get it back to Cartesian form. Attached my working for all 3.

Would anyone be able to help? And also, do my attempts for questions 2 & 3 look correct? Is 3 definitely just the use of harmonic formulae or is there more to it?

TIA
For 1) cant help think its a bit long.
* To get back to cartesian just multiply the mod by cos(arg) and sin(arg) to get the real and imaginary componets respectively. Its just polar form.
* To get the cube root, calc the mod arg form, then principle root is mod^(1/3) and arg/3. Then the other two root are same mod and arg+/- 2pi/3 .

2) Id expand the mulitple angles. So expand sin(2x) then factorise the sin part and reason about the cos -part (expand the cos(3x))

3) agree looks llike its harmonic form
(edited 1 month ago)
Original post by mqb2766
For 1) cant help think its a bit long.
* To get back to cartesian just multiply the mod by cos(arg) and sin(arg) to get the real and imaginary componets respectively. Its just polar form.
* To get the cube root, calc the mod arg form, then principle root is mod^(1/3) and arg/3. Then the other two root are same mod and arg+/- 2pi/3 .

2) Id partiallly do a mulitple angle. So expand sin(2x) then factorise sin and reason about the cos factor. Id not exxpand the cos(3x).

3) agree looks llike its harmonic form

Yeah I thought 1 was a bit long. I initially expanded (a+bi)^3 and tried comparing coefficients but it was so long so I gave up. I was also convinced I’d done it wrong when I did it your way as I ended up with decimals and I’m used to all the examples being nice neat fractions haha. Thank you, appreciate it.

Okay thanks re 2. I haven’t done trig for like 4 years so was just trying to reason it out bit by bit. Do my answers look alright for 2 & 3?
Original post by kwikmaffs
Yeah I thought 1 was a bit long. I initially expanded (a+bi)^3 and tried comparing coefficients but it was so long so I gave up. I was also convinced I’d done it wrong when I did it your way as I ended up with decimals and I’m used to all the examples being nice neat fractions haha. Thank you, appreciate it.

Somewhat to my surprise, it seems there's no general algebraic formula for the cube root of a complex number (I thought you must be able to do something with Cardano's formula, but if you try, you end up needing the cube root of other complex numbers - https://en.wikipedia.org/wiki/Casus_irreducibilis).
Original post by kwikmaffs
Yeah I thought 1 was a bit long. I initially expanded (a+bi)^3 and tried comparing coefficients but it was so long so I gave up. I was also convinced I’d done it wrong when I did it your way as I ended up with decimals and I’m used to all the examples being nice neat fractions haha. Thank you, appreciate it.

Okay thanks re 2. I haven’t done trig for like 4 years so was just trying to reason it out bit by bit. Do my answers look alright for 2 & 3?

Ive not worked them through and you can use wolfram to check the numbers if you want so
https://www.wolframalpha.com/input?i=sin%282x%29+%2B+sin%28x%29cos%283x%29%3D0
(obv its to be usesd senibly) but Id have spotted/dealt with the sin(x) = 0 solutions as you known the sin(2x) expansion and should "know" cos(3x). Then just concentrate on the (simple) cubic in cos(x). It shouldnt be more than a few lines.
Original post by kwikmaffs
Yeah I thought 1 was a bit long. I initially expanded (a+bi)^3 and tried comparing coefficients but it was so long so I gave up. I was also convinced I’d done it wrong when I did it your way as I ended up with decimals and I’m used to all the examples being nice neat fractions haha. Thank you, appreciate it.

Somewhat to my surprise, it seems there's no general algebraic formula for the cube root of a complex number (I thought you must be able to do something with Cardano's formula, but if you try, you end up needing the cube root of other complex numbers - https://en.wikipedia.org/wiki/Casus_irreducibilis).

So yeah, decimals, or a horrendous formula in terms of sin/cos of arctan(1/3)/3 are about the best you can do.
Original post by mqb2766
Ive not worked them through and you can use wolfram to check the numbers if you want so
https://www.wolframalpha.com/input?i=sin%282x%29+%2B+sin%28x%29cos%283x%29%3D0
(obv its to be usesd senibly) but Id have spotted/dealt with the sin(x) = 0 solutions as you known the sin(2x) expansion and should "know" cos(3x). Then just concentrate on the (simple) cubic in cos(x). It shouldnt be more than a few lines.

Okay thanks
Original post by DFranklin
Somewhat to my surprise, it seems there's no general algebraic formula for the cube root of a complex number (I thought you must be able to do something with Cardano's formula, but if you try, you end up needing the cube root of other complex numbers - https://en.wikipedia.org/wiki/Casus_irreducibilis).

So yeah, decimals, or a horrendous formula in terms of sin/cos of arctan(1/3)/3 are about the best you can do.

Okay thank you, appreciate it.
Original post by kwikmaffs
Okay thanks

For the harmonic form, it looks like youve misued the sin one, rather than cos as asked for in the question. If necessary, use the addition identity for cos and expand your answer to check (or chuck it into wolfram - its a nasty habit which is hard to give up).
(edited 1 month ago)
Original post by kwikmaffs
Got three questions to do, ~snip~

Just to say, another way to do question 2 is use the Sum<->Product identities: https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities

Another way that winds up feeling quite similar is to write $z = e^{ix}$ and then use $z^k+z^{-k}= 2 \cos kx, \quad z^k-z^{-k} = 2i \sin kx$

(I partially only mention this to try to provide a complex number based solution).
Original post by kwikmaffs
Got three questions to do, I’ve attempted them all and have an answer for 2 & 3 but no idea if they’re right. But I’m not sure how to continue question 1? I’ve found the polar form of 6+2j but not sure how to actually find the roots and get it back to Cartesian form. Attached my working for all 3.

Would anyone be able to help? And also, do my attempts for questions 2 & 3 look correct? Is 3 definitely just the use of harmonic formulae or is there more to it?

TIA

Note thinking abit more about the second one, a slightly ore elegant way would be to note
sin(x)cos(3x) = sin(4x)/2 - sin(2x)/2
(product to sum identity) which when combined with the first term gives you something like
sin(4x) = -sin(2x)
which you can simply take the arcsin of and reason about the multiple values. So simpler to treat in multple angle form, rather than expanding the multliple angles, though the latter is ore usual.
(edited 1 month ago)
Original post by kwikmaffs
Got three questions to do, I’ve attempted them all and have an answer for 2 & 3 but no idea if they’re right. But I’m not sure how to continue question 1? I’ve found the polar form of 6+2j but not sure how to actually find the roots and get it back to Cartesian form. Attached my working for all 3.

Would anyone be able to help? And also, do my attempts for questions 2 & 3 look correct? Is 3 definitely just the use of harmonic formulae or is there more to it?

TIA

You may want to check your solution to exercise 13 by plotting the graph of your answer and graph. Below is a plot of your solution and the question expression.