Inverse of complex number imaginary part
up vote
1
down vote
favorite
I have something like this
$$z+{frac{1}{i}}$$
Which I expand:
$$z+{frac{1}{i}} = x+yi+{frac{1}{i}} = x+i(y+?)$$
I'm not sure what should be in place of $?$.
Assuming $itimes x = {frac{1}{i}}$, then $x = {frac{1}{i}} div i = {frac{1}{i}} times {frac{1}{i}} = -1$. But definitely, $itimes (-1) neq {frac{1}{i}}$.
complex-numbers
asked Nov 25 at 19:09
user3132457
856
add a comment |
up vote
1
down vote
favorite
I have something like this
$$z+{frac{1}{i}}$$
Which I expand:
$$z+{frac{1}{i}} = x+yi+{frac{1}{i}} = x+i(y+?)$$
I'm not sure what should be in place of $?$.
Assuming $itimes x = {frac{1}{i}}$, then $x = {frac{1}{i}} div i = {frac{1}{i}} times {frac{1}{i}} = -1$. But definitely, $itimes (-1) neq {frac{1}{i}}$.
complex-numbers
asked Nov 25 at 19:09
user3132457
856
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have something like this
$$z+{frac{1}{i}}$$
Which I expand:
$$z+{frac{1}{i}} = x+yi+{frac{1}{i}} = x+i(y+?)$$
I'm not sure what should be in place of $?$.
Assuming $itimes x = {frac{1}{i}}$, then $x = {frac{1}{i}} div i = {frac{1}{i}} times {frac{1}{i}} = -1$. But definitely, $itimes (-1) neq {frac{1}{i}}$.
complex-numbers
asked Nov 25 at 19:09
user3132457
856
I have something like this
$$z+{frac{1}{i}}$$
Which I expand:
$$z+{frac{1}{i}} = x+yi+{frac{1}{i}} = x+i(y+?)$$
I'm not sure what should be in place of $?$.
Assuming $itimes x = {frac{1}{i}}$, then $x = {frac{1}{i}} div i = {frac{1}{i}} times {frac{1}{i}} = -1$. But definitely, $itimes (-1) neq {frac{1}{i}}$.
complex-numbers
complex-numbers
asked Nov 25 at 19:09
user3132457
856
asked Nov 25 at 19:09
user3132457
856
asked Nov 25 at 19:09
user3132457
856
asked Nov 25 at 19:09
user3132457
856
asked Nov 25 at 19:09
user3132457
856
856
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
We have
$$z+{frac{1}{i}} = x+yi+{frac{1cdot i}{icdot i}} = x+yi+{frac{i}{-1}}= x+iy-i=x+(y-1)i$$
and more in general for any $zin mathbb{C}, zneq 0$
$$frac1{z}=frac1{z}frac{bar z}{bar z}=frac{bar z}{z bar z}=frac{bar z}{mid z mid^2}$$
answered Nov 25 at 19:11
gimusi
90.6k74495
so, $${frac{1}{i}}=-i$$? I just don't understand how? Well by expanding, yes. But, by intuition, no.
– user3132457
Nov 26 at 3:45
@user3132457 Yes exactly and indeed, to confirm the result, we ca check that $-i$ is the inverse of $i$ indeed $$-icdot i=-(-1)=1$$.
– gimusi
Nov 26 at 6:24
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
We have
$$z+{frac{1}{i}} = x+yi+{frac{1cdot i}{icdot i}} = x+yi+{frac{i}{-1}}= x+iy-i=x+(y-1)i$$
and more in general for any $zin mathbb{C}, zneq 0$
$$frac1{z}=frac1{z}frac{bar z}{bar z}=frac{bar z}{z bar z}=frac{bar z}{mid z mid^2}$$
answered Nov 25 at 19:11
gimusi
90.6k74495
so, $${frac{1}{i}}=-i$$? I just don't understand how? Well by expanding, yes. But, by intuition, no.
– user3132457
Nov 26 at 3:45
@user3132457 Yes exactly and indeed, to confirm the result, we ca check that $-i$ is the inverse of $i$ indeed $$-icdot i=-(-1)=1$$.
– gimusi
Nov 26 at 6:24
add a comment |
up vote
0
down vote
accepted
We have
$$z+{frac{1}{i}} = x+yi+{frac{1cdot i}{icdot i}} = x+yi+{frac{i}{-1}}= x+iy-i=x+(y-1)i$$
and more in general for any $zin mathbb{C}, zneq 0$
$$frac1{z}=frac1{z}frac{bar z}{bar z}=frac{bar z}{z bar z}=frac{bar z}{mid z mid^2}$$
answered Nov 25 at 19:11
gimusi
90.6k74495
so, $${frac{1}{i}}=-i$$? I just don't understand how? Well by expanding, yes. But, by intuition, no.
– user3132457
Nov 26 at 3:45
@user3132457 Yes exactly and indeed, to confirm the result, we ca check that $-i$ is the inverse of $i$ indeed $$-icdot i=-(-1)=1$$.
– gimusi
Nov 26 at 6:24
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
We have
$$z+{frac{1}{i}} = x+yi+{frac{1cdot i}{icdot i}} = x+yi+{frac{i}{-1}}= x+iy-i=x+(y-1)i$$
and more in general for any $zin mathbb{C}, zneq 0$
$$frac1{z}=frac1{z}frac{bar z}{bar z}=frac{bar z}{z bar z}=frac{bar z}{mid z mid^2}$$
answered Nov 25 at 19:11
gimusi
90.6k74495
We have
$$z+{frac{1}{i}} = x+yi+{frac{1cdot i}{icdot i}} = x+yi+{frac{i}{-1}}= x+iy-i=x+(y-1)i$$
and more in general for any $zin mathbb{C}, zneq 0$
$$frac1{z}=frac1{z}frac{bar z}{bar z}=frac{bar z}{z bar z}=frac{bar z}{mid z mid^2}$$
answered Nov 25 at 19:11
gimusi
90.6k74495
answered Nov 25 at 19:11
gimusi
90.6k74495
answered Nov 25 at 19:11
gimusi
90.6k74495
answered Nov 25 at 19:11
gimusi
90.6k74495
90.6k74495
so, $${frac{1}{i}}=-i$$? I just don't understand how? Well by expanding, yes. But, by intuition, no.
– user3132457
Nov 26 at 3:45
@user3132457 Yes exactly and indeed, to confirm the result, we ca check that $-i$ is the inverse of $i$ indeed $$-icdot i=-(-1)=1$$.
– gimusi
Nov 26 at 6:24
add a comment |
so, $${frac{1}{i}}=-i$$? I just don't understand how? Well by expanding, yes. But, by intuition, no.
– user3132457
Nov 26 at 3:45
@user3132457 Yes exactly and indeed, to confirm the result, we ca check that $-i$ is the inverse of $i$ indeed $$-icdot i=-(-1)=1$$.
– gimusi
Nov 26 at 6:24
so, $${frac{1}{i}}=-i$$? I just don't understand how? Well by expanding, yes. But, by intuition, no.
– user3132457
Nov 26 at 3:45
so, $${frac{1}{i}}=-i$$? I just don't understand how? Well by expanding, yes. But, by intuition, no.
– user3132457
Nov 26 at 3:45
@user3132457 Yes exactly and indeed, to confirm the result, we ca check that $-i$ is the inverse of $i$ indeed $$-icdot i=-(-1)=1$$.
– gimusi
Nov 26 at 6:24
@user3132457 Yes exactly and indeed, to confirm the result, we ca check that $-i$ is the inverse of $i$ indeed $$-icdot i=-(-1)=1$$.
– gimusi
Nov 26 at 6:24
add a comment |
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