What is the factorization of $(da+dbi)^2$, $(da+dbi)^4$, and $(da+dbi)^8$,
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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
- $d1*a^2 + d2iab − 1*b^2$
$d1*a^2 + d^22iab − 1*b^2$
however, according to an online calculator $(da + dib)^2$ = $d^2a^2+2d^2abi−d^2b^2$
If $(a + bi)^4$ = $a^4 + 4a^3bi − 6a^2b^2 − 4ab^3i + b^4$
then what would:
$(da + dib)^4$ = ??
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: $(da + dib)^8$ = ??
complex-numbers factoring
asked Jan 13 at 23:53
Du'uzu MesDu'uzu Mes
135
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add a comment |
$begingroup$
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
- $d1*a^2 + d2iab − 1*b^2$
$d1*a^2 + d^22iab − 1*b^2$
however, according to an online calculator $(da + dib)^2$ = $d^2a^2+2d^2abi−d^2b^2$
If $(a + bi)^4$ = $a^4 + 4a^3bi − 6a^2b^2 − 4ab^3i + b^4$
then what would:
$(da + dib)^4$ = ??
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: $(da + dib)^8$ = ??
complex-numbers factoring
asked Jan 13 at 23:53
Du'uzu MesDu'uzu Mes
135
$endgroup$
add a comment |
$begingroup$
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
- $d1*a^2 + d2iab − 1*b^2$
$d1*a^2 + d^22iab − 1*b^2$
however, according to an online calculator $(da + dib)^2$ = $d^2a^2+2d^2abi−d^2b^2$
If $(a + bi)^4$ = $a^4 + 4a^3bi − 6a^2b^2 − 4ab^3i + b^4$
then what would:
$(da + dib)^4$ = ??
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: $(da + dib)^8$ = ??
complex-numbers factoring
asked Jan 13 at 23:53
Du'uzu MesDu'uzu Mes
135
$endgroup$
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
- $d1*a^2 + d2iab − 1*b^2$
$d1*a^2 + d^22iab − 1*b^2$
however, according to an online calculator $(da + dib)^2$ = $d^2a^2+2d^2abi−d^2b^2$
If $(a + bi)^4$ = $a^4 + 4a^3bi − 6a^2b^2 − 4ab^3i + b^4$
then what would:
$(da + dib)^4$ = ??
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: $(da + dib)^8$ = ??
complex-numbers factoring
complex-numbers factoring
asked Jan 13 at 23:53
Du'uzu MesDu'uzu Mes
135
asked Jan 13 at 23:53
Du'uzu MesDu'uzu Mes
135
edited Jan 14 at 0:01
Du'uzu Mes
asked Jan 13 at 23:53
Du'uzu MesDu'uzu Mes
135
asked Jan 13 at 23:53
Du'uzu MesDu'uzu Mes
135
asked Jan 13 at 23:53
Du'uzu MesDu'uzu Mes
135
135
add a comment |
add a comment |
1 Answer
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It looks like you simply need to factor out the variable $d$ from each expression. Note that
$$(da+dbi)^n = (d(a+bi))^n = d^n(a+bi)^n.$$
answered Jan 13 at 23:56
D.B.D.B.
1,61029
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1 Answer
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$begingroup$
It looks like you simply need to factor out the variable $d$ from each expression. Note that
$$(da+dbi)^n = (d(a+bi))^n = d^n(a+bi)^n.$$
answered Jan 13 at 23:56
D.B.D.B.
1,61029
$endgroup$
add a comment |
$begingroup$
It looks like you simply need to factor out the variable $d$ from each expression. Note that
$$(da+dbi)^n = (d(a+bi))^n = d^n(a+bi)^n.$$
answered Jan 13 at 23:56
D.B.D.B.
1,61029
$endgroup$
add a comment |
$begingroup$
It looks like you simply need to factor out the variable $d$ from each expression. Note that
$$(da+dbi)^n = (d(a+bi))^n = d^n(a+bi)^n.$$
answered Jan 13 at 23:56
D.B.D.B.
1,61029
$endgroup$
It looks like you simply need to factor out the variable $d$ from each expression. Note that
$$(da+dbi)^n = (d(a+bi))^n = d^n(a+bi)^n.$$
answered Jan 13 at 23:56
D.B.D.B.
1,61029
answered Jan 13 at 23:56
D.B.D.B.
1,61029
answered Jan 13 at 23:56
D.B.D.B.
1,61029
answered Jan 13 at 23:56
D.B.D.B.
1,61029
1,61029
add a comment |
add a comment |
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