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complex numbers confusion

So I've been trying to practice evaluating complex numbers raised to an integer power. Here's an example problem from a sheet that I followed and its solution:

https://cdn.discordapp.com/attachments/692344148212187226/1004881882187579413/Screenshot_39.png

However when I tried to evaluate (-1 + i)^7 using the same procedure from the solution above I got the wrong answer and the solution for it incorporated costheta + isintheta whereas the example from above didn't. Is there a problem with the example that I followed or have I done something wrong?

The video linked has the full solution to (-1 + i)^7:

https://youtu.be/hoOJb5oqSSk

Here's my full working (again imitating the steps used in the solution from the sheet):

https://media.discordapp.net/attachments/692344148212187226/1005077869313142804/20220805_123915.jpg?width=927&height=473
This is year 2 Further Maths.

21pi/4 = -3pi/4. This means your final answer should be 8root2 e^(-3pi/4), in exponential form, since 2^7/2 = 2^3 x 2^1/2 = 8root2. This is equivalent to -8 - 8i.
Original post by Driving_Mad
This is year 2 Further Maths.

21pi/4 = -3pi/4. This means your final answer should be 8root2 e^(-3pi/4), in exponential form, since 2^7/2 = 2^3 x 2^1/2 = 8root2. This is equivalent to -8 - 8i.


Ah I see, so the method from the sheet gets you directly to the modulus of the final complex number. Also where did the negative in front of 3pi/4 come from and why was 21pi/4 divided by 7?
Reply 3
Original post by Driving_Mad
This is year 2 Further Maths.

21pi/4 = -3pi/4. This means your final answer should be 8root2 e^(-3pi/4), in exponential form, since 2^7/2 = 2^3 x 2^1/2 = 8root2. This is equivalent to -8 - 8i.

I presume you're working mod 2pi. What you've written isn't exactly right, is it? :biggrin:
Original post by davros
I presume you're working mod 2pi. What you've written isn't exactly right, is it? :biggrin:


could you elaborate please?
Original post by Soul Wavel3ngth
could you elaborate please?

Think of an Argand diagram. The real axis starts at argument 0. You go 2pi radians around you end up in the same position you started in. Going around 2pi again gives an argument of 4pi which ends up at the positive real axis again, so 0 is equivalent to 2kpi (k is an integer).

This means if you have 21pi/4 you can keep subtracting 2pi until the argument ends up in the principal form between -pi and pi.

Original post by davros
I presume you're working mod 2pi. What you've written isn't exactly right, is it? :biggrin:

Should have made it clearer in my post :smile:
Reply 6
Original post by Soul Wavel3ngth
could you elaborate please?


In addition to Driving_Mad's update. note that sin and cos are periodic with period 2pi, so you can add/subtract multiples of 2pi from any value you have and still get the same answer when you take the sine or cosine, so e.g. sin(21pi/4) = sin(21pi/4 - 2pi) = sin(21pi/4 - 4pi) etc
...or, in short, what goes round comes round :wink:
(edited 1 year ago)
Original post by davros
In addition to Driving_Mad's update. note that sin and cos are periodic with period 2pi, so you can add/subtract multiples of 2pi from any value you have and still get the same answer when you take the sine or cosine, so e.g. sin(21pi/4) = sin(21pi/4 - 2pi) = sin(21pi/4 - 4pi) etc


It feels it would have been a lot clearer if the given solution had decomposed 250iπ250i\pi as 125(2iπ)125(2i\pi) instead of 2(125iπ)2(125i\pi).

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