complex num q. (1 Viewer)

[shkspeare]

hi anyone able to help me with this

Let z = a+ ib where a² + b² doesnt equal 0

Show that if Im(z) < 0 then Im(1/z) > 0

thanks

 

[wogboy]

z = a + ib
Im(z) = b

1/z = 1/(a + ib)
= (a - ib)/(a + ib)(a - ib)
= (a - ib)/(a^2 + b^2)
= a/(a^2 + b^2) - ib/(a^2 + b^2)

therefore,
Im(1/z) = -b/(a^2 + b^2)

since Im(z) < 0, then b < 0.

now since a^2 + b^2 > 0 for all real a,b (given that a^2 + b^2 =/= 0),

-b/(a^2 + b^2) > 0

therefore Im(1/z) > 0.

 

[shkspeare]

ooooOo thx a lot

 

[spice girl]

alternatively

let z = Rcis@ (@ = theta)

then 1/z = (1/R)cis(-@) by de-moivre's thm

Im(z) = Rsin@
if Im(z) < 0 then sin@ < 0
then sin(-@) > 0 (sin is an odd function)
then (1/R)sin@ > 0

 

[mazza_728]

IM SO LOST! I havent started 4 unit - my teacher has been realy unorganised - so would that explain things ? or is everyone suppose to know this mumbojumbo?

 

[McLake]

Originally posted by mazza_728
IM SO LOST! I havent started 4 unit - my teacher has been realy unorganised - so would that explain things ? or is everyone suppose to know this mumbojumbo?

If you havn't started yet then don't worry too much, but here is a brief explianation of the solution:

z = a + ib <- generic complex number representation.
Im(z) = b <- defintion

1/z = 1/(a + ib) <- follows logically
= (a - ib)/(a + ib)(a - ib) <- a trick known as "realising" the denomenator, simlar to rationlisation for square roots as denomenators
= (a - ib)/(a^2 + b^2) <- since i^2 = -1, -i^2 = 1
= a/(a^2 + b^2) - ib/(a^2 + b^2)

therefore,
Im(1/z) = -b/(a^2 + b^2) <- defintion

The rest is logic

since Im(z) < 0, then b < 0.

now since a^2 + b^2 > 0 for all real a,b (given that a^2 + b^2 =/= 0),

-b/(a^2 + b^2) > 0

therefore Im(1/z) > 0.