Equivalence of a complex number

Question

We have the complex number: $$ z = \frac{(1+i)^n}{(1-i)^{n-2}}$$ Having $n \in Z, n\ge 2$

I am asked for obtaining an equivalent of $z$ and selecting one of the provided 4 options.

Once I computed several algebraic steps I reached the outcome: $-2i^{n+1}$

The four options are:

a) $-2i$ if n is even and $-2$ if n is odd.

b) $-2(-1)^ni$.

c) $-2i(-1)^{n/2}$ is n is even and $2(-1)^{\frac{n-1}{2}}$ if n is odd.

d) $2(-1)^{n+1}$

My issue is that I do not know how to interpret $-2i^{n+1}$ in order to select one of these 4. Could you help me out?

Answer 1

We have that

$$z = \frac{(1+i)^n}{(1-i)^{n-2}}= (1+i)^2\frac{(1+i)^{n-2}}{(1-i)^{n-2}}= (2i)\left(\frac{1+i}{1-i}\frac{1+i}{1+i}\right)^{n-2}= (2i)\left(\frac{2i}{2}\right)^{n-2}=2(i)^{n-1}$$

and $2(i)^{n-1}$is equal to $-2i(-1)^{n/2}$ is n is even and $2(-1)^{\frac{n-1}{2}}$ if n is odd, indeed

• for $n=3 \implies 2(i)^{2}=-2$
• for $n=5 \implies 2(i)^{4}=2$
• for $n=7 \implies 2(i)^{6}=-2$
• ...

and

• for $n=2 \implies 2(i)=2i$
• for $n=4 \implies 2(i)^{3}=-2i$
• for $n=6 \implies 2(i)^{5}=2i$
• ...

or as an alternative

• for $n$ even $2i^{n-1}=2i^{2k-1} = 2i(-1)^{k-1}$

and

• for $n$ odd $2i^{n-1}=2i^{2k} = 2(-1)^k$

Answer 2

Observe that

$$\frac{1+i}{1-i}=i\implies\frac{(1+i)^n}{(1-i)^{n-2}}=\left(\frac{1+i}{1-i}\right)^{n-2}(1+i)^2=i^{n-2}2i\ldots$$

Answer 3

Since $\displaystyle{1\over 1-i} = {1+i\over 2}$ we have:

$${(1+i)^n\over (1-i)^{n-2}} = {(1+i)^{2n-2}\over 2^{n-2}}={(2i)^{n-1}\over 2^{n-2}}=2i^{n-1}=...$$

Answer 4

Upon doing the following operation $$\frac{(1+i)}{(1-i)}= \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i - 1}{1 + 1} = i$$ So $$\frac{(1+i)^n}{(1-i)^{n-2}} = \frac{(1+i)^{n-2}}{(1-i)^{n-2}}(1+i)^2 = i^{n-2}(1+i)^2 = i^{n-2}(1 + 2i - 1) = i^{n-2}(2i) = 2i^{n-1}$$