Proof for arguments of complex numbers (1 Viewer)

[ashokkumar]

I was just wondering, is the arg(1/z) = arg(conjugate of z)?

If so, how can you prove this?

 

[adomad]

by looking at an argand diagram, , if you plot the point x+iy to be some where in the first quad, the point x-iy will have the same side ratios and hence magnitude of the angle is the same, but it is just flipped into the 4th quad... hope this helps

 

[ashokkumar]

by looking at an argand diagram, , if you plot the point x+iy to be some where in the first quad, the point x-iy will have the same side ratios and hence magnitude of the angle is the same, but it is just flipped into the 4th quad... hope this helps

hmmm.. i understood everything you did, except the last part where you said -arg(z) = arg(conjugate z)

 

[adomad]

hmmm.. i understood everything you did, except the last part where you said -arg(z) = arg(conjugate z)

that bit comes from looking at the diagram... ill draw one up.. where x-iy is the conjugate of x+iy

 

[ashokkumar]

oooh. the diagram shows that the arg(z) = arg(conjugate z). So how does that show that -arg(z) = arg(conjugate of z)?

is it because arg(z) = -arg(z)

 

[adomad]

oooh. the diagram shows that the arg(z) = arg(conjugate z). So how does that show that -arg(z) = arg(conjugate of z)?

is it because arg(z) = -arg(z)

they are equal in magnitude, but the sign of the arg, or angle, for the conjugate is minus, because it is below the x-axis.... ill give u an example.






hence, from this wonderful example, we can conclude that arg(z) X-1 = arg(conjugate of z)

 

[pman]

oooh. the diagram shows that the arg(z) = arg(conjugate z). So how does that show that -arg(z) = arg(conjugate of z)?

is it because arg(z) = -arg(z)

what disgram are you looking at dude..pretty sure it shows that arg(z)=-arg(conjugate z)

 

[ashokkumar]

oooh i got it

thanks guys