Fast way to solve $(s-2i)^2 (s+2i)^2$
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2
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$(s-2i)^2 (s+2i)^2=$
$(s^2-4si-4) (s^2+4si-4)=$
$s^4+4s^3i-4s^2-4s^3i+16s^2+16si-4s^2-16si+16 =$
$(s^4-8s^2+16) =$
$(s^2+4)^2$
Is there a quicker way to see that $(s-2i)^2 (s+2i)^2= (s^2+4)^2$
algebra-precalculus complex-numbers factoring
asked Nov 17 at 21:40
roy212
184
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up vote
2
down vote
favorite
$(s-2i)^2 (s+2i)^2=$
$(s^2-4si-4) (s^2+4si-4)=$
$s^4+4s^3i-4s^2-4s^3i+16s^2+16si-4s^2-16si+16 =$
$(s^4-8s^2+16) =$
$(s^2+4)^2$
Is there a quicker way to see that $(s-2i)^2 (s+2i)^2= (s^2+4)^2$
algebra-precalculus complex-numbers factoring
asked Nov 17 at 21:40
roy212
184
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$(s-2i)^2 (s+2i)^2=$
$(s^2-4si-4) (s^2+4si-4)=$
$s^4+4s^3i-4s^2-4s^3i+16s^2+16si-4s^2-16si+16 =$
$(s^4-8s^2+16) =$
$(s^2+4)^2$
Is there a quicker way to see that $(s-2i)^2 (s+2i)^2= (s^2+4)^2$
algebra-precalculus complex-numbers factoring
asked Nov 17 at 21:40
roy212
184
$(s-2i)^2 (s+2i)^2=$
$(s^2-4si-4) (s^2+4si-4)=$
$s^4+4s^3i-4s^2-4s^3i+16s^2+16si-4s^2-16si+16 =$
$(s^4-8s^2+16) =$
$(s^2+4)^2$
Is there a quicker way to see that $(s-2i)^2 (s+2i)^2= (s^2+4)^2$
algebra-precalculus complex-numbers factoring
algebra-precalculus complex-numbers factoring
asked Nov 17 at 21:40
roy212
184
asked Nov 17 at 21:40
roy212
184
asked Nov 17 at 21:40
roy212
184
asked Nov 17 at 21:40
roy212
184
asked Nov 17 at 21:40
roy212
184
184
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
We have
$$(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$$
and recall that $(A-B)(A+B)=A^2-B^2$.
answered Nov 17 at 21:41
gimusi
88.7k74394
I don't see why $(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$
– roy212
Nov 17 at 21:46
@roy212 We have in general (not only for complex number) $$A^2cdot B^2 =(Acdot B)^2$$
– gimusi
Nov 17 at 21:47
And $A^2*B^2*C^2*D^2 = (A*B*C*D)^2$?
– roy212
Nov 17 at 21:56
@roy212 Yes of course! It is also true.
– gimusi
Nov 17 at 21:57
As long as you are dealing with commutative things. So, okay for real and complex numbers but not quaternions or matrices.
– badjohn
Nov 18 at 15:57
add a comment |
up vote
1
down vote
$$(s-2i)^2(s+2i)^2=left[(s-2i)(s+2i)right]^2=left[s^2-4i^2right]^2=ldots$$
answered Nov 17 at 21:41
DonAntonio
176k1491224
add a comment |
up vote
0
down vote
notice:
$$(a+b)(a-b)=a^2-b^2$$
so:
$$(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$$
answered Nov 17 at 21:47
Henry Lee
1,676118
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We have
$$(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$$
and recall that $(A-B)(A+B)=A^2-B^2$.
answered Nov 17 at 21:41
gimusi
88.7k74394
I don't see why $(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$
– roy212
Nov 17 at 21:46
@roy212 We have in general (not only for complex number) $$A^2cdot B^2 =(Acdot B)^2$$
– gimusi
Nov 17 at 21:47
And $A^2*B^2*C^2*D^2 = (A*B*C*D)^2$?
– roy212
Nov 17 at 21:56
@roy212 Yes of course! It is also true.
– gimusi
Nov 17 at 21:57
As long as you are dealing with commutative things. So, okay for real and complex numbers but not quaternions or matrices.
– badjohn
Nov 18 at 15:57
add a comment |
up vote
3
down vote
accepted
We have
$$(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$$
and recall that $(A-B)(A+B)=A^2-B^2$.
answered Nov 17 at 21:41
gimusi
88.7k74394
I don't see why $(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$
– roy212
Nov 17 at 21:46
@roy212 We have in general (not only for complex number) $$A^2cdot B^2 =(Acdot B)^2$$
– gimusi
Nov 17 at 21:47
And $A^2*B^2*C^2*D^2 = (A*B*C*D)^2$?
– roy212
Nov 17 at 21:56
@roy212 Yes of course! It is also true.
– gimusi
Nov 17 at 21:57
As long as you are dealing with commutative things. So, okay for real and complex numbers but not quaternions or matrices.
– badjohn
Nov 18 at 15:57
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We have
$$(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$$
and recall that $(A-B)(A+B)=A^2-B^2$.
answered Nov 17 at 21:41
gimusi
88.7k74394
We have
$$(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$$
and recall that $(A-B)(A+B)=A^2-B^2$.
answered Nov 17 at 21:41
gimusi
88.7k74394
answered Nov 17 at 21:41
gimusi
88.7k74394
answered Nov 17 at 21:41
gimusi
88.7k74394
answered Nov 17 at 21:41
gimusi
88.7k74394
88.7k74394
I don't see why $(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$
– roy212
Nov 17 at 21:46
@roy212 We have in general (not only for complex number) $$A^2cdot B^2 =(Acdot B)^2$$
– gimusi
Nov 17 at 21:47
And $A^2*B^2*C^2*D^2 = (A*B*C*D)^2$?
– roy212
Nov 17 at 21:56
@roy212 Yes of course! It is also true.
– gimusi
Nov 17 at 21:57
As long as you are dealing with commutative things. So, okay for real and complex numbers but not quaternions or matrices.
– badjohn
Nov 18 at 15:57
add a comment |
I don't see why $(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$
– roy212
Nov 17 at 21:46
@roy212 We have in general (not only for complex number) $$A^2cdot B^2 =(Acdot B)^2$$
– gimusi
Nov 17 at 21:47
And $A^2*B^2*C^2*D^2 = (A*B*C*D)^2$?
– roy212
Nov 17 at 21:56
@roy212 Yes of course! It is also true.
– gimusi
Nov 17 at 21:57
As long as you are dealing with commutative things. So, okay for real and complex numbers but not quaternions or matrices.
– badjohn
Nov 18 at 15:57
I don't see why $(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$
– roy212
Nov 17 at 21:46
I don't see why $(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$
– roy212
Nov 17 at 21:46
@roy212 We have in general (not only for complex number) $$A^2cdot B^2 =(Acdot B)^2$$
– gimusi
Nov 17 at 21:47
@roy212 We have in general (not only for complex number) $$A^2cdot B^2 =(Acdot B)^2$$
– gimusi
Nov 17 at 21:47
And $A^2*B^2*C^2*D^2 = (A*B*C*D)^2$?
– roy212
Nov 17 at 21:56
And $A^2*B^2*C^2*D^2 = (A*B*C*D)^2$?
– roy212
Nov 17 at 21:56
@roy212 Yes of course! It is also true.
– gimusi
Nov 17 at 21:57
@roy212 Yes of course! It is also true.
– gimusi
Nov 17 at 21:57
As long as you are dealing with commutative things. So, okay for real and complex numbers but not quaternions or matrices.
– badjohn
Nov 18 at 15:57
As long as you are dealing with commutative things. So, okay for real and complex numbers but not quaternions or matrices.
– badjohn
Nov 18 at 15:57
add a comment |
up vote
1
down vote
$$(s-2i)^2(s+2i)^2=left[(s-2i)(s+2i)right]^2=left[s^2-4i^2right]^2=ldots$$
answered Nov 17 at 21:41
DonAntonio
176k1491224
add a comment |
up vote
1
down vote
$$(s-2i)^2(s+2i)^2=left[(s-2i)(s+2i)right]^2=left[s^2-4i^2right]^2=ldots$$
answered Nov 17 at 21:41
DonAntonio
176k1491224
add a comment |
up vote
1
down vote
up vote
1
down vote
$$(s-2i)^2(s+2i)^2=left[(s-2i)(s+2i)right]^2=left[s^2-4i^2right]^2=ldots$$
answered Nov 17 at 21:41
DonAntonio
176k1491224
$$(s-2i)^2(s+2i)^2=left[(s-2i)(s+2i)right]^2=left[s^2-4i^2right]^2=ldots$$
answered Nov 17 at 21:41
DonAntonio
176k1491224
answered Nov 17 at 21:41
DonAntonio
176k1491224
answered Nov 17 at 21:41
DonAntonio
176k1491224
answered Nov 17 at 21:41
DonAntonio
176k1491224
176k1491224
add a comment |
add a comment |
up vote
0
down vote
notice:
$$(a+b)(a-b)=a^2-b^2$$
so:
$$(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$$
answered Nov 17 at 21:47
Henry Lee
1,676118
add a comment |
up vote
0
down vote
notice:
$$(a+b)(a-b)=a^2-b^2$$
so:
$$(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$$
answered Nov 17 at 21:47
Henry Lee
1,676118
add a comment |
up vote
0
down vote
up vote
0
down vote
notice:
$$(a+b)(a-b)=a^2-b^2$$
so:
$$(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$$
answered Nov 17 at 21:47
Henry Lee
1,676118
notice:
$$(a+b)(a-b)=a^2-b^2$$
so:
$$(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$$
answered Nov 17 at 21:47
Henry Lee
1,676118
answered Nov 17 at 21:47
Henry Lee
1,676118
answered Nov 17 at 21:47
Henry Lee
1,676118
answered Nov 17 at 21:47
Henry Lee
1,676118
1,676118
add a comment |
add a comment |
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