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Relations And Functions

Question
CBSEENMA11012957

Convert each of the complex numbers given in the Polar form.

space space space space space space space space space space space space space square root of 3 plus space straight i

Solution

Here,    z = space space space space space space space square root of 3 space end root plus space straight i
Let       z =      space space space space space space space space square root of 3 plus space straight i space equals space straight r space left parenthesis cos space straight theta space plus space straight i space sin space straight theta right parenthesis                                   ...(1)
space space space space space space space space space space space space space space rightwards double arrow     space space space space space space space space space space space space space space space space straight r space cos space straight theta space equals space square root of 3                                                   ...(ii)
and                    space space space space space space space space space space space space space space space straight r space sin space straight theta space equals space 1                                                   ...(iii)

Squaring and adding (ii) and (iii), we get
  WiredFaculty
  rightwards double arrow            r2 = 4              rightwards double arrow              r = 2
From (ii),          space space space space space space space space space space space space space space space space space space 2 space cos space straight theta space.. square root of 3 space space space space space space space space space space space space space rightwards double arrow space space space space space space space cos space straight theta space space equals space fraction numerator square root of 3 over denominator 2 end fraction
From (iii),              WiredFaculty
since both space space space space space sin space straight theta space and space cos space straight theta are positive

∴     space space space theta lies in 1st quadrant

∴     WiredFaculty


Putting in (i), we get  space space space space space space space space space space space space space space space space space space space space space space space space straight z space equals space 2 open parentheses cos straight pi over 6 plus straight i space sin straight pi over 6 close parentheses
Which is the required polar form.
Hence,          space space space space space space space space space space space space space space space open vertical bar straight z close vertical bar space equals space space straight r equals space 2 space semicolon space arg space left parenthesis straight z right parenthesis space equals space straight theta space equals space straight pi over 6