Complex Number Raised to a Power - Identity Proof

Question

I was wondering if anyone can help explain the following identity/equation to me

(z+i)^4 = 1 + i

The context for the problem is shown in the picture below, and is featured on a 2008 AMC math competition problem set.

A solution is offered however this equation is a crucial component in it. Any help would be much appreciated. Thanks!

Answer 1

Hint:  by binomial expansion: $$\require{cancel}(z+i)^4 = z^4 + 4 i z^3 + 6 i^2z^2 + 4 i^3 z + i^4 = \cancel{(z^4 + 4 i z^3 - 6 z^2 - 4 i z \color{red}{-i})} + 1 + \color{red}{i} = 1+i$$

Answer 2

First, recognize that $(z+i)^4=z^4+4z^3i-6z^2-4zi+1$ as per the binomial theorem.

For reference, the binomial theorem states that for complex $x,y$ and natural $n$ one has

$$(x+y)^n = \sum\limits_{k=0}^n \binom{n}{k}x^ky^{n-k}$$

Next, given the equation in the question... we were told that $z^4+4z^3i-6z^2-4zi-i=0$

So, we have $(z+i)^4-1-i=z^4+4z^3i-6z^2-4zi+1-1-i=z^4+4z^3i-6z^2-4zi-i=0$

which implies that $(z+i)^4=1+i$

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Note: this identity is true given the conditions in the question. For arbitrary values of $z$ which do not necessarily follow the conditions given in the question, we of course do not necessarily have $(z+i)^4=1+i$. For trivial counter example, consider $z=-i$ in which case we have $(z+i)^4=0^4=0\neq 1+i$

Answer 3

Some things to think about:

The roots of an $z^n = 1$ from a regular polygon with n-sides. The roots all lie on a circle of radius 1.

The roots of an $z^n = k^n$ lie on a circle of radius $|k|.$

$(z-\omega)^n = k^n$ re-centers that polygon but does not change the area.

$(z-\omega)^n = k^n(\cos \theta + i\sin \theta)$ rotates the polygon but does not change the area.

Answer 4

It's not an identity. It's an equation to be solved.

$z^4+z^3i-6z^2-4z-i=0$

$z^4+4z^3i-6z^2-4z+1=1+i\iff $

$z^4+4z^3i+6z^2i^2+4zi^3+i^4=1+i \iff $

$(z+i)^4 =1+i $

So it's only true for $z $ where $z^4+z^3i-6z^2-4z-i=0$. It'll be false for all other $z $.

Answer 5

$$(z+i)^4=z^4+4z^{3}i+6z^{2}i^2+4zi^3+i^4=z^4+4iz^3-6z^2-4iz+1=f(z)+(1+i)$$