# Complex number calculation and expresing it in polar form

• December 15th 2011, 08:38 AM
calmath
Complex 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, 08:53 AM
Plato
Re: Complex number calculation and expresing it in polar form
Quote:

Originally Posted by calmath
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

That is not the way it is done here.
You show some work. Tell us what you don't understand.

Here is a start: $1-i=\sqrt{2}\exp\left(\frac{-\pi}{4}\right).$
• December 15th 2011, 04:54 PM
calmath
Re: Complex number calculation and expresing it in polar form
Quote:

Originally Posted by Plato
That is not the way it is done here.
You show some work. Tell us what you don't understand.

Here is a start: $1-i=\sqrt{2}\exp\left(\frac{-\pi}{4}\right).$

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, 05:17 PM
Prove It
Re: Complex number calculation and expresing it in polar form
Quote:

Originally Posted by calmath
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

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, 03:19 AM
FernandoRevilla
Re: Complex number calculation and expresing it in polar form
Out of curiosity: for a complex number $z=x+yj$ with $|x|=|y|$ the computation of $(x\pm yj)^n$ in binomial form can be simplified taking into account that $(x\pm yj)^2$ is a pure imaginary number. In our case

$(1-j)^5=[(1-j)^2]^2(1-j)=(-2j)^2(1-j)=-4(1-j)=-4+4j$ .