Use DeMoivre's Theorem to find the power of the complex number : (1-i)^5 (write solution in a+bi form)

Question

Complex numbers in polar form

Answer 1

Hi Cheyenne,

 

De Moivre's Theorem states that for any complex number (cosx + isinx),

 

(cosx + isinx)ⁿ = cos(nx) + isin(nx)

 

Also, remember Euler's identity, e^ix = cosx + isinx  

 

Can we write z = 1 - i in polar form, z = re^iθ?

 

Yes. If a complex number z = x + iy, then the modulus r = (x² + y²)^0.5 and the phase θ is given by arctan(y/x).

 

By inspection we see that x = 1 and y = -1. Then, r = [(1)² + (-1)²]^0.5 = √2

 

θ = arctan(-1/1) = -π/4

 

We also know that if z = re^iθ then z = r(cosθ + isinθ)

 

So, by DeMoivre's Theorem, z⁵ =  [r(cosθ + isinθ)]⁵ = = r⁵(cosθ + isinθ)⁵ = r⁵[cos(5θ) + isin(5θ)], where z = 1 - i.

 

Recall, r = √2 and θ = -π/4. So, (1 - i)⁵ = (√2)⁵[cos(-5π/4) + isin(-5π/4)] = 2^5/2[-√2/2 + i√2/2) = 2^5/2√2/2(-1 + i) = 2^6/2 / 2(-1 + i) = 2³/2(-1 + i) = (8/2)(-1 + i) = 4(-1 + i)