let u = 3-i
|u| = √(9+1) = √10
tanθ = -1/3
θ = 5.96143.. since θ in in quad IV in the Argand plane
u = √10(cos 5.96143.. + i sin 5.96143)
then z = √u = √√10(cos 2.980717... + i sin 2.980717...) by De Moivre
= -1.7553 + .28485 i <----- the primary square root of 3-i
by adding π to 2.980717.. you can find a 2nd square root
ii) w = az just extends the maginitude of z, it does not change θ
iii) use De Moivre
Let z= √(3 - i)
i) Plot z on an Argand diagram.
ii) Let w = az where a > 0, a E R. Express w in polar form
iii) Express w^8 in the form ka^n(x + i√y) where k,x,y E Z
2 answers
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