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.

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- December 15th 2011, 07:38 AMcalmathComplex number calculation and expresing it in polar form
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. - December 15th 2011, 07:53 AMPlatoRe: Complex number calculation and expresing it in polar form
- December 15th 2011, 03:54 PMcalmathRe: Complex number calculation and expresing it in polar form
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 - December 15th 2011, 04:17 PMProve ItRe: Complex number calculation and expresing it in polar form
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. - December 16th 2011, 02:19 AMFernandoRevillaRe: Complex number calculation and expresing it in polar form
Out of curiosity: for a complex number with the computation of in binomial form can be simplified taking into account that is a pure imaginary number. In our case

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