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Lecture 6

MATH 1C03 Lecture Notes - Lecture 6: Rational Number


Department
Mathematics
Course Code
MATH 1C03
Professor
Miroslav Lovric
Lecture
6

Page:
of 3
:ture6
Example
÷
=-3 -Si Convert
to
polar form z=r( cosotisinotrciso
ring --HPtfs5M=My
a- aretantta )-arctanl
'S)returns avalue between -%andTY2
-counterclockwise
0=1.03 so ,-3 -5i=34h' cis (4.17)
-5 .-
in our case ,not unique 󲍻cis( 4.17+2 't ')
-3 1.03 F
,),}A- 1.03+7=4.17 If K= -1 ,
OR -3 -Si .
.Mc ;s( -2.11 )youget
-clockwise -
.--5
Example
÷
mpvlethepo1arformof3-3ir-fEH-T-3M2oarctanHH.Ft2tkWhyPo1a3-3i-3M2.cisfTYtt2tHZi-rilcos9tisinAi3Zz-rzkosO2tisinOd1.Ie.3.4Z.5krolcos9tisinAHcoso2tisinod-rir2.cosQcosos-sino.sinoDComputingPoTlRktilsin.osinsotcosacosoDDefinitioniei0-cosotisino@tampabay.rr.com
,+q ).
÷irrational z,
Tryo :# number
ei "= costtisint .-1.10=-1 o
,+!
iZ2
o°°e#H=O
Homework
÷
Iirtslcos1oioDtisinCa-o.t7rkosotisinOHz2-r2kos2otisin2ad-r2cis2o7n-rhCcosnotisinno17DeMoivrisformulaitzirei07zn.rneintlrei9n-rn.eioyn-rh@CiQneinoEx_plendzttitz-3-3i-z-3M2cisf7pEEl3M2Yt.cistFtlorcisY7j7I-3RttasHt-ltfT-f2aiftxAt0dhfm.uk
.im#=29t+3ft=3Fe=Ef
xample
÷
3=1 ,z= ?x3=1x=3M=/
This istheonlyreal solution
z=P .c
;s(5
Angle -3=0
(osA=CosB
\
Too Zrcisa .?23=1 A=B2tK
<
Hoo "r3¢os3O+isin30)
O=0t2Tk
.v=L .kosotisino )O=2±i
󲍻r3=1󲍻r=1 3
2=1 .cis (01=1 k=O
7=1
.ci#t=k.'z+i)k=lz=l.cislt5t)=-t2-iE2k=2