Complex numbers, find (3/(1-i)-(1-i)/2)^40?

[iamavisitor]

Homework Statement

Please help me find (3/(1-i)-(1-i)/2)^40. I got a result (see below) but I'm not sure whether it is correct. Any help is appreciated. Thanks.

Homework Equations

The Attempt at a Solution

I got (1+2i)^40. After this I got some funny numbers like 7^10*16*(6232-474*i).

 

[Mark44]

Homework Statement

Please help me find (3/(1-i)-(1-i)/2)^40. I got a result (see below) but I'm not sure whether it is correct. Any help is appreciated. Thanks.

Homework Equations

The Attempt at a Solution

I got (1+2i)^40.

I get this (above) also.

After this I got some funny numbers like 7^10*16*(6232-474*i).

The way to go here is to convert 1 + 2i into polar form, using the Theorem of de Moivre. Then raising to the 40th power is easy.
See http://en.wikipedia.org/wiki/De_Moivre's_formula.

 

[iamavisitor]

I know about de Moivre's formula. With it I get radius (r) to be sqrt(5) and angle 63.43 degrees. Again this is complicated to do by hand and I am wandering whether I got something wrong.

 

[ehild]

It is correct. But it might be that you or somebody else copied the problem wrong. (3/(1-i)-(1+i)/2)^40 would be easy.

ehild

 

[Mark44]

I know about de Moivre's formula. With it I get radius (r) to be sqrt(5) and angle 63.43 degrees. Again this is complicated to do by hand and I am wandering whether I got something wrong.

No, it's not complicated to do by hand once you have the complex number in this form.

[r*e^iθ]ⁿ = rⁿ*e^niθ