Calculate (1-j)^5 and express the answer in polar form. 'j' being the imaginary number.
If you could also show me the working out i'd appreciate it.
I got up to the following part;
Where it says calculate I didnt quite understand what it wants so id did the following;
(1-i)^5=1-i-i^5=2-i
Modulus is;
Sqrt(2^2+1^2 = Sqrt(5)
Argument(2-i)= atan(-1/2)= -26.6
(atan(-1/2))+360= 333.4degrees
In polar form, 2-i is: Sqrt(5), theta= 333.4
Wrong. If you were going to expand these brackets, you should have (1 - i)(1 - i)(1 - i)(1 - i)(1 - i), which should give you a LOT more than 1 - i - i^5.
As you have been told already, convert to the polar form which Plato did in post 2, then use DeMoivre's Theorem, like you have been instructed.
Out of curiosity: for a complex number $\displaystyle z=x+yj$ with $\displaystyle |x|=|y|$ the computation of $\displaystyle (x\pm yj)^n$ in binomial form can be simplified taking into account that $\displaystyle (x\pm yj)^2$ is a pure imaginary number. In our case
$\displaystyle (1-j)^5=[(1-j)^2]^2(1-j)=(-2j)^2(1-j)=-4(1-j)=-4+4j$ .