Finding magnitude of a complex number
up vote
7
down vote
favorite
$$z = dfrac{2+2i}{4-2i}$$
$$|z| = ? $$
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
36.4k114591
asked Dec 3 at 17:47
Hamilton
1828
add a comment |
up vote
7
down vote
favorite
$$z = dfrac{2+2i}{4-2i}$$
$$|z| = ? $$
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
36.4k114591
asked Dec 3 at 17:47
Hamilton
1828
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
$$z = dfrac{2+2i}{4-2i}$$
$$|z| = ? $$
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
36.4k114591
asked Dec 3 at 17:47
Hamilton
1828
$$z = dfrac{2+2i}{4-2i}$$
$$|z| = ? $$
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
36.4k114591
asked Dec 3 at 17:47
Hamilton
1828
edited Dec 4 at 13:46
greedoid
36.4k114591
asked Dec 3 at 17:47
Hamilton
1828
edited Dec 4 at 13:46
greedoid
36.4k114591
edited Dec 4 at 13:46
greedoid
36.4k114591
edited Dec 4 at 13:46
greedoid
36.4k114591
36.4k114591
asked Dec 3 at 17:47
Hamilton
1828
asked Dec 3 at 17:47
Hamilton
1828
asked Dec 3 at 17:47
Hamilton
1828
1828
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
7
down vote
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
36.4k114591
add a comment |
up vote
3
down vote
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
221k15155273
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
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
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
36.4k114591
add a comment |
up vote
7
down vote
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
36.4k114591
add a comment |
up vote
7
down vote
up vote
7
down vote
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
36.4k114591
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
36.4k114591
answered Dec 3 at 17:49
greedoid
36.4k114591
answered Dec 3 at 17:49
greedoid
36.4k114591
answered Dec 3 at 17:49
greedoid
36.4k114591
36.4k114591
add a comment |
add a comment |
up vote
3
down vote
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
221k15155273
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
up vote
3
down vote
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
221k15155273
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
up vote
3
down vote
up vote
3
down vote
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
221k15155273
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
221k15155273
answered Dec 3 at 17:51
lab bhattacharjee
221k15155273
answered Dec 3 at 17:51
lab bhattacharjee
221k15155273
answered Dec 3 at 17:51
lab bhattacharjee
221k15155273
221k15155273
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
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024418%2ffinding-magnitude-of-a-complex-number%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
