What is the factorization of $(da+dbi)^2$, $(da+dbi)^4$, and $(da+dbi)^8$,

Question

I need a clarification for factorization when the variable 'd' is included in $(a + ib)^n$, such that $(da + dib)^n$.

Specifically for three cases.

If $(a + ib)^2$ = $1*a^2 + 2iab − 1*b^2$, then what would: $(da + dib)^2$ = ??

For the first one I thought that $(da + dib)^2$ was either

1. $d1*a^2 + d2iab − 1*b^2$

2. $d1*a^2 + d^22iab − 1*b^2$

   however, according to an online calculator $(da + dib)^2$ = $d^2a^2+2d^2abi−d^2b^2$

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If $(a + bi)^4$ = $a^4 + 4a^3bi − 6a^2b^2 − 4ab^3i + b^4$

then what would: $(da + dib)^4$ = ??

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If $(a + bi)^8$ = $a^8 + 8a^7bi − 28a^6b^2 − 56a^5b^3i + 70a^4b^4 + 56a^3b^5i − 28a^2b^6 − 8ab^7i + b^8$

then what would: $(da + dib)^8$ = ??

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Answer 1

It looks like you simply need to factor out the variable $d$ from each expression. Note that $$(da+dbi)^n = (d(a+bi))^n = d^n(a+bi)^n.$$