Complex Numbers 1. if Sol:
= 4 determine the locus of z. let z = x1 + iy 1
(x1 - !" + (y 1 + 1!" = 1# x1" - #x1 + $ + y 1" + "y 1 + 1 - 1# = %. x1" + y 1" - #x1 + "y 1 - # = % ... &e'uired locus is x" + y " - #x + "y - # = % ". f z = " - i) then sho* tht z" - 4z + 1 = % Sol: Consider z = " - i =, z - " = - i S'urin on both sides *e et (z - "!" = (-i!" z" - 4z + 4 = $i" z" - 4z + 4 = -$ (... i" = -1! z" - 4z + 1 = %. . ind the the multiplicti/e multiplicti/e in/erse of 0 + "4i Sol: he multiplicti/e multiplicti/e in/erse of of + ib is
con2ute of z 1 is z" 3. ind the s'ure root of ( + 4i! sol: S'ure root of + ib
Comprin rel prts *e et x=
4x " = 1
"x = 1
4x" - 1 = % 0. xpress the complex number into modulus - mplitude form) z = - 1 - i Sol:
5i/en tht z = - 1 - i 6et z = x + iy Comprin *e et x = - 1) y = -
7e 8no* tht x = r cos ) y = r sin
.. .
cos nd sin re neti/e) the re'uired nle lies in the third 'udrnt) so nle is neti/e.
he mplitude of complex number is 8no*n s rument denoted by 9r (z! = 9r (x + iy! = tn-1 ( ! 9r ( ! = 9r (x - iy! 9r (z1. z"! = 9r z1 + 9r z " + n ) n -1) %) 1; 9r (
! = 9r z - 9r z + n 1 "
)n
-1) %) 1;
he sin of rument chnes dependin on the 'udrnts ccordinly. is re'uired rument.
<. f the 9r ( ! nd 9r ( Sol:
! re
respecti/ely) find (9r z1 + 9r z "!
6et z1 = x1 - i y 1) =, = x1 + iy 1 So the point lies in the 'udrnt
De Moiver's Theorem
3. Sho* tht (>1 + i ! n + (>1 > i ! n = (>1!n "n+1 cos n? . Sol.
[email protected] = (>1 + i ! n + (>1 > i ! n
= A>1 (1 > i !B n + A>1 (1 + i !B n = (>1!n A(1 > i ! n + (1 + i !nB 6et us find the od - mplitude form of 1 + i 6et 1 + i = x + iy. x = 1 y =
.
