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MSSIGNMENli 2
~
r r r
litI: QUE s- ID NS FIllI:l:.Yj AN WERED r
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DUE: II JUNE 2D26 === +
DISCLAIMER: I I
- TRJS- DOCUMENT I DR7l"EFERENCE7'lnrGUT ~ cPURP 5£
ACADEMIC MISCONDUCT RESULTING FROM THEIR USE. IT IS YDURi
RESPONSIBILITY T ENSU E ORIGINALITY AN CDMPUIAN E WITH
l ---
ONLY. I DD NDTtTAKE RESPPNSIBIUITY !FOR A~Y PLAGIARl~M. tMISUSE. OR
'RELEVANT GUI ELINES DR STANDARDS. ~
, PLEASE USE THIS DOCUMENT AS A GUIDE ONLY
SECTION A: MULTIPLE CHOICE
1. Please insert and complete the following table to answer the questions in Section A:
Question 1(a) 1(b) 1(c) 1(d)
Answers iii iv ii ii
SECTION B: LONGER ANSWERS
2. To prove an "if and only if" (⟺) statement, we must show that the implication holds in both
directions, or use a chain of reversible algebraic steps. Note that for the expression to be defined,
the denominator cannot be zero, which directly gives the restriction z =−3.
Step 1: Simplify the expression by multiplying by the complex conjugate
Let z=x+iy. We substitute this into the rational function and multiply the numerator and denominator
by the complex conjugate of the denominator to make it real-valued.
; -3 - (z --3) +·iy
"~" '"
-
z + 3 (:c +·.3) +iy
"
The complex conjugate of the denominator is (x+3)−iy. Multiplying both top and bottom:
z-3 l(z-3) +ir)[(z.+3}-; WJ
z + s:= l(:z + 3) + ~11ll(!t + sJ - iyJ
Step 2: Expand the numerator and denominator
Denominator:
, Numerator:
(x - 3)(x + 3) - i(x - 3)y + i(x + 3)y - (iy) 2
2
- (x - 9) - ixy + 3iy + ixy + 3iy + y 2
= (:z:2 + y 2 - 9) + i(6y)
Putting it together:
2 2
z- 3 x +y - 9
- - - - - - - - +i
.( 6y )
z +3 (:z: + 3)2 + y2 (:z: + 3)2 ,... y2
Step 3: Extract the Real Part and set it to 0
z- 3 :z:2 + y2 - 9
Re
z+ 3 (:z: + 3)2 + y2
For the real part to equal 0:
x2 + y2 - 9 - 0
(x + 3)2 -1- y2
----+-----+---+ MlT3 7D5 ~
MSSIGNMENli 2
~
r r r
litI: QUE s- ID NS FIllI:l:.Yj AN WERED r
r
-+-----+----
DUE: II JUNE 2D26 === +
DISCLAIMER: I I
- TRJS- DOCUMENT I DR7l"EFERENCE7'lnrGUT ~ cPURP 5£
ACADEMIC MISCONDUCT RESULTING FROM THEIR USE. IT IS YDURi
RESPONSIBILITY T ENSU E ORIGINALITY AN CDMPUIAN E WITH
l ---
ONLY. I DD NDTtTAKE RESPPNSIBIUITY !FOR A~Y PLAGIARl~M. tMISUSE. OR
'RELEVANT GUI ELINES DR STANDARDS. ~
, PLEASE USE THIS DOCUMENT AS A GUIDE ONLY
SECTION A: MULTIPLE CHOICE
1. Please insert and complete the following table to answer the questions in Section A:
Question 1(a) 1(b) 1(c) 1(d)
Answers iii iv ii ii
SECTION B: LONGER ANSWERS
2. To prove an "if and only if" (⟺) statement, we must show that the implication holds in both
directions, or use a chain of reversible algebraic steps. Note that for the expression to be defined,
the denominator cannot be zero, which directly gives the restriction z =−3.
Step 1: Simplify the expression by multiplying by the complex conjugate
Let z=x+iy. We substitute this into the rational function and multiply the numerator and denominator
by the complex conjugate of the denominator to make it real-valued.
; -3 - (z --3) +·iy
"~" '"
-
z + 3 (:c +·.3) +iy
"
The complex conjugate of the denominator is (x+3)−iy. Multiplying both top and bottom:
z-3 l(z-3) +ir)[(z.+3}-; WJ
z + s:= l(:z + 3) + ~11ll(!t + sJ - iyJ
Step 2: Expand the numerator and denominator
Denominator:
, Numerator:
(x - 3)(x + 3) - i(x - 3)y + i(x + 3)y - (iy) 2
2
- (x - 9) - ixy + 3iy + ixy + 3iy + y 2
= (:z:2 + y 2 - 9) + i(6y)
Putting it together:
2 2
z- 3 x +y - 9
- - - - - - - - +i
.( 6y )
z +3 (:z: + 3)2 + y2 (:z: + 3)2 ,... y2
Step 3: Extract the Real Part and set it to 0
z- 3 :z:2 + y2 - 9
Re
z+ 3 (:z: + 3)2 + y2
For the real part to equal 0:
x2 + y2 - 9 - 0
(x + 3)2 -1- y2
