Finding magnitude of a complex number
$$z = dfrac{2+2i}{4-2i}$$
$$|z| = ? $$
My attempt:
$$dfrac{(2+2i)(4+2i)}{(4-2i)(4+2i)} = dfrac{4+12i}{20} = dfrac{4}{20}+dfrac{12}{20}i = dfrac{1}{5} + dfrac{3}{5}i$$
Now taking its magnitude and we have that
$$|z| = sqrt{biggr (dfrac 1 5 biggr ) ^2 +biggr (dfrac 3 5 biggr )^2} = sqrt {dfrac 2 5 }$$
Am I right?
complex-numbers
edited Dec 4 at 13:46
greedoid
37.2k114794
asked Dec 3 at 17:47
Hamilton
1838
add a comment |
$$z = dfrac{2+2i}{4-2i}$$
$$|z| = ? $$
My attempt:
$$dfrac{(2+2i)(4+2i)}{(4-2i)(4+2i)} = dfrac{4+12i}{20} = dfrac{4}{20}+dfrac{12}{20}i = dfrac{1}{5} + dfrac{3}{5}i$$
Now taking its magnitude and we have that
$$|z| = sqrt{biggr (dfrac 1 5 biggr ) ^2 +biggr (dfrac 3 5 biggr )^2} = sqrt {dfrac 2 5 }$$
Am I right?
complex-numbers
edited Dec 4 at 13:46
greedoid
37.2k114794
asked Dec 3 at 17:47
Hamilton
1838
add a comment |
$$z = dfrac{2+2i}{4-2i}$$
$$|z| = ? $$
My attempt:
$$dfrac{(2+2i)(4+2i)}{(4-2i)(4+2i)} = dfrac{4+12i}{20} = dfrac{4}{20}+dfrac{12}{20}i = dfrac{1}{5} + dfrac{3}{5}i$$
Now taking its magnitude and we have that
$$|z| = sqrt{biggr (dfrac 1 5 biggr ) ^2 +biggr (dfrac 3 5 biggr )^2} = sqrt {dfrac 2 5 }$$
Am I right?
complex-numbers
edited Dec 4 at 13:46
greedoid
37.2k114794
asked Dec 3 at 17:47
Hamilton
1838
$$z = dfrac{2+2i}{4-2i}$$
$$|z| = ? $$
My attempt:
$$dfrac{(2+2i)(4+2i)}{(4-2i)(4+2i)} = dfrac{4+12i}{20} = dfrac{4}{20}+dfrac{12}{20}i = dfrac{1}{5} + dfrac{3}{5}i$$
Now taking its magnitude and we have that
$$|z| = sqrt{biggr (dfrac 1 5 biggr ) ^2 +biggr (dfrac 3 5 biggr )^2} = sqrt {dfrac 2 5 }$$
Am I right?
complex-numbers
complex-numbers
edited Dec 4 at 13:46
greedoid
37.2k114794
asked Dec 3 at 17:47
Hamilton
1838
edited Dec 4 at 13:46
greedoid
37.2k114794
asked Dec 3 at 17:47
Hamilton
1838
edited Dec 4 at 13:46
greedoid
37.2k114794
edited Dec 4 at 13:46
greedoid
37.2k114794
edited Dec 4 at 13:46
greedoid
37.2k114794
37.2k114794
asked Dec 3 at 17:47
Hamilton
1838
asked Dec 3 at 17:47
Hamilton
1838
asked Dec 3 at 17:47
Hamilton
1838
1838
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Yes, you are. You can do it also like this: $$Big|{2+2iover 4-2i}Big|=Big|{1+iover 2-i}Big|={|1+i|over |2-i|}= {sqrt{2}over sqrt{5}}$$
answered Dec 3 at 17:49
greedoid
37.2k114794
add a comment |
It is better to use $$left|dfrac abright|=dfrac{|a|}{|b|}$$
$|2+2i|=sqrt{2^2+2^2}=2sqrt2$
answered Dec 3 at 17:51
lab bhattacharjee
222k15155273
I first made the denominator real and then did my calculations to avoid any errors .
– Hamilton
Dec 3 at 17:55
@Hamilton, At least for modulus, the formula I believe is more suitable
– lab bhattacharjee
Dec 3 at 17:56
However, is it a bad way?
– Hamilton
Dec 3 at 18:11
Rationalization may be costly in some cases
– lab bhattacharjee
Dec 3 at 18:17
@Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
– timtfj
Dec 3 at 19:53
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes, you are. You can do it also like this: $$Big|{2+2iover 4-2i}Big|=Big|{1+iover 2-i}Big|={|1+i|over |2-i|}= {sqrt{2}over sqrt{5}}$$
answered Dec 3 at 17:49
greedoid
37.2k114794
add a comment |
Yes, you are. You can do it also like this: $$Big|{2+2iover 4-2i}Big|=Big|{1+iover 2-i}Big|={|1+i|over |2-i|}= {sqrt{2}over sqrt{5}}$$
answered Dec 3 at 17:49
greedoid
37.2k114794
add a comment |
Yes, you are. You can do it also like this: $$Big|{2+2iover 4-2i}Big|=Big|{1+iover 2-i}Big|={|1+i|over |2-i|}= {sqrt{2}over sqrt{5}}$$
answered Dec 3 at 17:49
greedoid
37.2k114794
Yes, you are. You can do it also like this: $$Big|{2+2iover 4-2i}Big|=Big|{1+iover 2-i}Big|={|1+i|over |2-i|}= {sqrt{2}over sqrt{5}}$$
answered Dec 3 at 17:49
greedoid
37.2k114794
answered Dec 3 at 17:49
greedoid
37.2k114794
answered Dec 3 at 17:49
greedoid
37.2k114794
answered Dec 3 at 17:49
greedoid
37.2k114794
37.2k114794
add a comment |
add a comment |
It is better to use $$left|dfrac abright|=dfrac{|a|}{|b|}$$
$|2+2i|=sqrt{2^2+2^2}=2sqrt2$
answered Dec 3 at 17:51
lab bhattacharjee
222k15155273
I first made the denominator real and then did my calculations to avoid any errors .
– Hamilton
Dec 3 at 17:55
@Hamilton, At least for modulus, the formula I believe is more suitable
– lab bhattacharjee
Dec 3 at 17:56
However, is it a bad way?
– Hamilton
Dec 3 at 18:11
Rationalization may be costly in some cases
– lab bhattacharjee
Dec 3 at 18:17
@Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
– timtfj
Dec 3 at 19:53
|
show 1 more comment
It is better to use $$left|dfrac abright|=dfrac{|a|}{|b|}$$
$|2+2i|=sqrt{2^2+2^2}=2sqrt2$
answered Dec 3 at 17:51
lab bhattacharjee
222k15155273
I first made the denominator real and then did my calculations to avoid any errors .
– Hamilton
Dec 3 at 17:55
@Hamilton, At least for modulus, the formula I believe is more suitable
– lab bhattacharjee
Dec 3 at 17:56
However, is it a bad way?
– Hamilton
Dec 3 at 18:11
Rationalization may be costly in some cases
– lab bhattacharjee
Dec 3 at 18:17
@Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
– timtfj
Dec 3 at 19:53
|
show 1 more comment
It is better to use $$left|dfrac abright|=dfrac{|a|}{|b|}$$
$|2+2i|=sqrt{2^2+2^2}=2sqrt2$
answered Dec 3 at 17:51
lab bhattacharjee
222k15155273
It is better to use $$left|dfrac abright|=dfrac{|a|}{|b|}$$
$|2+2i|=sqrt{2^2+2^2}=2sqrt2$
answered Dec 3 at 17:51
lab bhattacharjee
222k15155273
answered Dec 3 at 17:51
lab bhattacharjee
222k15155273
answered Dec 3 at 17:51
lab bhattacharjee
222k15155273
answered Dec 3 at 17:51
lab bhattacharjee
222k15155273
222k15155273
I first made the denominator real and then did my calculations to avoid any errors .
– Hamilton
Dec 3 at 17:55
@Hamilton, At least for modulus, the formula I believe is more suitable
– lab bhattacharjee
Dec 3 at 17:56
However, is it a bad way?
– Hamilton
Dec 3 at 18:11
Rationalization may be costly in some cases
– lab bhattacharjee
Dec 3 at 18:17
@Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
– timtfj
Dec 3 at 19:53
|
show 1 more comment
I first made the denominator real and then did my calculations to avoid any errors .
– Hamilton
Dec 3 at 17:55
@Hamilton, At least for modulus, the formula I believe is more suitable
– lab bhattacharjee
Dec 3 at 17:56
However, is it a bad way?
– Hamilton
Dec 3 at 18:11
Rationalization may be costly in some cases
– lab bhattacharjee
Dec 3 at 18:17
@Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
– timtfj
Dec 3 at 19:53
I first made the denominator real and then did my calculations to avoid any errors .
– Hamilton
Dec 3 at 17:55
I first made the denominator real and then did my calculations to avoid any errors .
– Hamilton
Dec 3 at 17:55
@Hamilton, At least for modulus, the formula I believe is more suitable
– lab bhattacharjee
Dec 3 at 17:56
@Hamilton, At least for modulus, the formula I believe is more suitable
– lab bhattacharjee
Dec 3 at 17:56
However, is it a bad way?
– Hamilton
Dec 3 at 18:11
However, is it a bad way?
– Hamilton
Dec 3 at 18:11
Rationalization may be costly in some cases
– lab bhattacharjee
Dec 3 at 18:17
Rationalization may be costly in some cases
– lab bhattacharjee
Dec 3 at 18:17
@Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
– timtfj
Dec 3 at 19:53
@Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
– timtfj
Dec 3 at 19:53
|
show 1 more comment
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