Proving Properties of z When n is Odd/Even Integer

[maobadi]

Homework Statement

if z = 1/sqrt 2 + i/sqrt2

Homework Equations

show that z + z⁴ⁿ⁺¹ = 0,
when n is any odd integer.

if n is an even integer, find z + z⁴ⁿ⁺¹ in rectangular form.

The Attempt at a Solution

I have no clue where to start...

 

[ravioli]

What's z in polar form? Also note that $$\text{z} + \text{z} ^ \text{4n+1} = \text{z} * ( 1 + \text{z} ^ \text{4n} )=0$$

 

[Architeuthis Dux]

Hello! I can provide some guidance on how to approach this problem.

First, let's rewrite z in rectangular form: z = (1+i)/sqrt(2) = 1/sqrt(2) + i/sqrt(2).

Now, we can use the properties of complex numbers to simplify our expressions. Remember that (a+bi)(c+di) = (ac-bd) + (ad+bc)i.

For the first part of the problem, we want to show that z + z4n+1 = 0 when n is an odd integer. To do this, we can plug in our expression for z and simplify:

z + z4n+1 = (1/sqrt(2) + i/sqrt(2)) + (1/sqrt(2) + i/sqrt(2))4n+1

= (1/sqrt(2) + i/sqrt(2)) + (1/sqrt(2) + i/sqrt(2))(4n+1)

= (1/sqrt(2) + i/sqrt(2)) + (1/sqrt(2) + i/sqrt(2))(4n) + (1/sqrt(2) + i/sqrt(2))(1)

= (1/sqrt(2) + i/sqrt(2)) + (1/sqrt(2) + i/sqrt(2)) + (1/sqrt(2) + i/sqrt(2))

= 3(1/sqrt(2) + i/sqrt(2))

= 3z

Since n is an odd integer, we know that 4n+1 is also an odd integer. Therefore, we can rewrite our expression as:

z + z4n+1 = 3z = 3(1/sqrt(2) + i/sqrt(2))

= (3/sqrt(2)) + (3i/sqrt(2))

= (3/sqrt(2)) + i(3/sqrt(2))

= (3/sqrt(2))(1+i)

= (3/sqrt(2))(sqrt(2)/sqrt(2))

= 3sqrt(2)/2

= 0

Therefore, we have shown that z + z4n+1 = 0 when n is an odd integer