Powers of a complex number (1 Viewer)

[mazza_728]

Hi guys
as some as of you may know im trying to complete the first topic - complex numbers - these holidays because we havent actually started the 4unit topic at my school as yet! ive been going along ok except, with a few minor hiccups but now ive come across a huge one! Powers of a complex number, i understand how to do this and everything but i keep getting things like cos1+isin1 or cos-1+isin-2 or cos-1/2+isin-1/2 or cos-4/3+isin-4/3 or cos4+isin4. when i put cos-1 in my calculator i get .9998, the answer in the book is a whole number, do they round this up or is there another way of solving these equations.
This is my working for one of the questions:
(2+2i)^3
Let z=2+2i
=2(sqrt)2(cos1+isin1)
Hence z^3=16(sqrt)2(cos3+isin3)
=????
Can someone please help
thankyou xoxo

 

[ND]

Originally posted by mazza_728
for one of the questions:
(2+2i)^3
Let z=2+2i
=2(sqrt)2(cos1+isin1)

To get the argument of z, you need to take the arctan of 2/2=1. Which is pi/4.

 

[KeypadSDM]

Originally posted by mazza_728
when i put cos-1 in my calculator i get .9998

Try putting it into radian mode. In complex no's only radian mode is used.

 

[mazza_728]

its still not working in radian mode i get random numbers and ND i dont know wat ur talking about.. .can u explain?

 

[Constip8edSkunk]

when converting the cartesian form into r(cos@+isin@) format, tan@ = y/x. So, in the example above tan@ = 1 and @ =pi/4 (NOT 1).

.'.2+2i = 2sqrt[2] * [cos pi/4 + isin pi/4].

then use the de moirves th. to find the powers

 

[mazza_728]

Ok comprehende thankyou guys! i did this for all the others just must of freaked out and forgotten .. thankyou
happy new years xoxo

 

[Xayma]

Originally posted by KeypadSDM
In complex no's only radian mode is used.

But you can convert that to degrees for ease of calculation as long as you convert back for the answer.

 

[ND]

Originally posted by Xayma
But you can convert that to degrees for ease of calculation as long as you convert back for the answer.

Don't bother; there's just no need. Always work in radians (except in bearings).

 

[mazza_728]

ok ive got another one i cant do (1-i)^3(2+2i)^3
i think im making mistakes in my working but i keep seeming to make the same mistakes?? please help

 

[wogboy]

The key thing here is to follow the steps carefully & accurately.

1. Convert the complex number from rectangular form (a + ib), into polar form (r cis@). e.g. 1 + i = sqrt(2) * cis(pi/4)

2. Use De Moivre's theorem (rcis@)^n = r^n * cis(n@)

3. Convert back to rectangular form.

--------------------------------------------

(1 - i)^3 * (2+2i)^3
= [(1-i)(2+2i)]^3
= (2 + 2i - 2i + 2)^3
= 4^3
= 64

For the question above you didn't need to use De Moivre's Theorem at all. If you wanted to you could find (1-i)^3 and (2+2i)^3 separately (using De Moivre's Theorem) and then multiply together, but this would be longer & messy.

 

[Grey Council]

LOL, i didnt even notice you could make:
(1 - i)^3 * (2+2i)^3
into
= [(1-i)(2+2i)]^3

lol, i did it the long way. The :
If you wanted to you could find (1-i)^3 and (2+2i)^3 separately (using De Moivre's Theorem) and then multiply together
way. aaahhh, sometimes im so stupid.

 

[mazza_728]

thats wat i got except i think it was -64 somehow! the working and everything looks great but the answer is suppose to be 128+128i