A seemingly easy question...help me (1 Viewer)

[speedie]

z=cos(theta)+isin(theta)

express 1+z in mod/arg form

 

[speedie]

maybe it will help with the next part:
Hence show that (1+z)^4= 16(cos theta/2)^4 x (cis2theta)

 

[zeek]

1) 1+z = 1+0i+z
= cis(0) + cos(@) + isin(@)
= 2cos²(@/2) + isin(@)
im not sure how to change it from here into cis form. Once you get it into cis form, use it in ur next question

 

[Elvin]

z+1=cos@ + isin@ + 1 = cos@ + 1 + isin@
-> by using cos2@=2cos^2(@) - 1
cos@=2cos^2(@/2) - 1
cos@ + 1 =2cos^2(@/2)
so..z+1= 2cos^2(@/2) + isin@
--> by using sin2@=2sin@cos@
sin@=2sin(@/2)cos(@/2)
so...z+1= 2cos^2(@/2) + 2isin(@/2)cos(@/2)
take out 2cos(@/2) as a common factor
z+1=2cos^2(@/2) [cis@/2]
then (z+1)^4=blablabla

 

[speedie]

1) 1+z = 1+0i+z
= cis(0) + cos(@) + isin(@)
= 2cos²(@/2) + isin(@)
im not sure how to change it from here into cis form. Once you get it into cis form, use it in ur next question

zeek, how did cis(0)+cos(@) become 2cos^2(@/2) ???

 

[speedie]

ok nevermind i got it... wow this is a tough question

i did it a geometric way, which was a bit less confusing

 

[Affinity]

the argand plane is just there for you to visualise and gain insight, doesn't remove the need to blast through the algebra