) z r θ ( ) ( ) ( ) = then. Complete Solutions to Examination Questions Complete Solutions to Examination Questions 10.
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1 Complete Solutions to Examination Questions 0 Complete Solutions to Examination Questions 0. (i We need to determine + given + j, j: + + j + j (ii The product ( ( + j6 + 6 j 8 + j is given by ( + j( j j + j6 j Using FOIL [ ] + j + Because j + j. (a We need to find given + j, j: ( + j( j 6 j8 + j j Using FOIL [ ] 6 j5 + Because j 0 j5 (b How do we find the square root of + j? We first write + j in polar form: + j + tan / 5 ( ( 6.56 / The square root of 5 ( 6.56 is given by De-Moivre s Theorem (if n n r ( nθ : / / / / 5 ( 6.56 ( r θ ( 60 The other root has the same modulus but we add 80 because 80 5 ( ( 9.8 / The two roots of are 5.8 and (a We need to write ( ( then j in polar form and then take the fourth root of this: / j ( + ( tan / ( ( Remember j is in the third quadrant and that is why we had to add 80. What do we need to find?
2 Complete Solutions to Examination Questions 0 The four roots of this number. Let j ( be the first root 0 / 0 ( 60 Because ( One of the roots is ( 60. How many more roots do we have? Three more. How do we find these? By adding , , each time to the argument of ( ( ( ( ( ( ( ( ( 50, 0 and 0 Plotting these on the Argand diagram: Im Re (b We need to place each of these numbers into polar form: cos( 5 + j sin ( 5 ( 5 cos( + j sin ( ( Evaluating the given expression
3 Complete Solutions to Examination Questions 0 ( cos5 j sin 5 ( 5 cos + j sin ( Our final answer is 5. + ( ( 5 ( ( n n [ r θ ] r ( nθ Using r θ r ( ( 5 Using ( θ β q β q 5. (a We need to solve the quadratic equation x 6x+ 0. How? Use the quadratic formula with a, b 6 and c : ( ( ( b± b ac 6 ± 6 x a 6± 6 6 ± j ± j Our roots with imaginary number i j we have + i, i. (b The exponential form of a complex number is iθ e cos θ + isin θ ( ( π With θ we have iπ / π π e cos + isin + i ( + i i (c De Moivre s theorem says that if e θ then n inθ e cos( nθ + isin ( nθ We have n inθ + e + n inθ e θ θ e + e Using the rules of indices a n a cos( nθ + isin ( nθ + cos( nθ isin ( nθ The value of a is. in in n cos ( nθ inθ e
4 Complete Solutions to Examination Questions 0 5. We use the complex numbers a + i, b i, c + i: (a (i Evaluating ( a+ b c ( + i+ i( + i ( 5 i( + i 5 + 0i+ i 8i [ Using FOIL] 5+ i+ 8 + i (ii The ( bc * means determine the complex conjugate of bc. We first find bc: bc i + i ( ( + 8i+ i i 0+ i The complex conjugate of bc 0 + i is ( bc * 0 i (iii Remember to divide two complex numbers we have to multiple the numerator and denominator by the complex conjugate of the denominator. b i ( i( i a + i + 6 i 9i+ i i 0..i 0 0 (iv What does the notation ab mean? The modulus of ab. We first find the product ab and then the modulus. ab ( + i( i 6 9i + i i 9 7i (v What does the notation Re( ac mean? 9 7i Find the real part of the product ac. We have ac + i + i ( ( + i i+ i 7+ i The real part is equal to 7. (vi Need to convert each of the given complex numbers into polar form: a + i + 0 arg( a tan 0.8 Therefore a in polar form is 0 ( 0.8. Similarly we have b i + (, arg ( b tan ( ( ( c + i + 7, arg c tan.58 + π.858 We need to add π radians to the argument of c because the calculator gives the wrong c quadrant. We have b ( and ( Therefore
5 Complete Solutions to Examination Questions 0 5 ac b ( ( ( ( ( ( The modulus is.66 and argument.0. (b To apply De Moivre s theorem we first have convert the given complex number into polar form. We can find the argument in degrees so that we get a feel for the sie of the angle. i + ( tan 0 8. ( What does De Moivre s theorem state? n n If r θ then r ( nθ. We have / ( i 0 ( 8. / / ( 0 8. ( / / ( /6 m ( 6.5 Because ( a a One of the roots of ( i / is 0 /6 ( 6.5 r. How many roots are there? Three roots because we are asked to find the cube roots. How do we find the other two roots? 60 Add 0 each time. Let r and be the other two roots: r /6 /6 r ( ( ( ( r + /6 / We first convert 8i into polar form. On the Argand diagram 8i is plotted as: 0 Im n mn Re
6 Complete Solutions to Examination Questions 0 6 Clearly by examining this diagram we have 8i 8 ( 90 given by taking the number to the index /. Let. The cube roots of this number are be the first root then / / 8 ( ( We have two other roots and. Where are these? These have a modulus but the argument of : 60 0 is added each time to and then to ( ( 50 and ( ( The three roots of 8i are (, ( and ( We need to find the 5 th roots of because we are given polar form? On the Argand diagram lies 5. What is in y x has a modulus of and argument of 80, that is 80 We can find one of the roots of one of the roots then ( ( by taking this ( 80 /5 to the index /5. Let be 80 /5 n n 80 ( 6 Because r ( nθ 5 The other four roots are found by adding 60 7 to each root which results in 5 + ( 6 7 ( 08 ( ( 80 ( ( 5 5 ( ( πi / 8. (a We need to convert w e into rectangular form: πi / π π w e cos + isin + i [ + i] Taking out The product w is given by
7 Complete Solutions to Examination Questions 0 7 w ( + 5i ( + i ( + i 5i+ 5i ( ( 8 i ( + i Because 8 i ( +i ( + i / Because / The quotient is given by w + 5i ( + 5i( i Complex Conjugate of w ( i ( i is i i 5i 5i ( i 8i 8i / [ i] Because / The addition + w is equal to + w ( + 5i + ( + i i i We now place each of these results in polar form: w ( + i ( i + tan ( π ( Since w is in the third quadrant that is why we have added π radians. The quotient result [ i] 7 ( tan ( w (.58 The addition + w in polar form is + w i ( (b Let a+ bi where a and b are real. Substituting this into the given equation + yields: ( a+ bi + ( a bi Remember if a+ ib then a ib ( [ ] Expanding a abi bi a bi ( ( a b + a+ ab b i Remember bi i b b
8 Complete Solutions to Examination Questions 0 8 Equating real and imaginary parts gives a b + a and ab b 0 b a 0 From the last equation b( a ( 0 we have b 0 or a. If b 0 then the first equation a b + a becomes a + a which implies a + a+ 0. This quadratic equation gives us complex roots but we are assuming that a is real. Hence b cannot equal ero, that is b 0 which means that a. Substituting a into a b + a gives b + 6 b b 6 ± Our values are a+ bi + 6 i, 6i. (c In polar form we have + i ( + tan π 0 How do we find? Use De Moivre s Theorem which states that if r θ n n r nθ : 0 then ( 0 π 0 0π π 0 Remember 0 π π π + radians is π radians because π radians is a complete revolution which means we are back at our starting point. π 0π
9 Complete Solutions to Examination Questions 0 9 What does / mean? The cube roots of. How many are there? Three. We can find the first root by applying De Moivre s Theorem: / / π / π / π 9 We have two more roots and the other roots have exactly the same modulus but argument is increased by π each time: / π π / 7π / 7π π / π ( ( cos( θ + isin ( θ cos ( θ + cos ( θ isin ( θ + cos( θ i sin ( θ + i sin ( θ (d Using the hint means we need to expand cos( θ isin ( θ +. Apply Binomial expansion: ( θ i ( θ ( θ ( θ ( θ i ( θ ( θ ( θ ( θ i ( θ ( θ ( θ ( cos + cos sin cos sin sin + Collecting real and imaginary terms in the last line. cos θ + i sin θ θ. Raising this to the index gives cos cos sin cos sin sin We can write ( ( cos( θ + i sin ( θ ( θ ( θ ( θ + i ( θ ( θ + i ( θ cos sin cos sin (* Equating the real and imaginary parts of ( and (* gives cos θ cos θ cos θ sin θ ( ( ( ( 9. Need to be careful with this question because of minus sign in front of the 0. How can we write the minus 0 in polar form? Remember 0 on the Argand diagram lies π Therefore 0 in polar form is 0 iπ iπ iπ 0e 0e e i e π /0 /0 i π π 0 i 9π 0 because modulus is 0 and argument is π. We have 0e Applying the rules of indices a a a 0e The argument of the complex number is 9 π. 0 m n m+ n
10 Complete Solutions to Examination Questions (a We are given the complex numbers j and w + j. We need to determine, + w, w,, w What does denote? The conjugate of which means that + j. Adding the two complex numbers gives + w j + + j 5+ j ( ( The product w can be found by using FOIL: w j + j 6+ j j j ( ( 6+ j+ 8+ j The reciprocal /w is given by j j j w + j + What does denote? This is the modulus of. We have j + 5 (b Need to write j in modulus argument form: ( ( arg arg tan π ( ( j π + Remember j is in the third quadrant therefore we have added π radians. in modulus argument form is We are told that w π. j therefore w ( j /. We apply De Moivre s theorem to find the square root of w. De Moivre s theorem states that if r θ then r nθ. Let be one of the roots then using the polar form of we have n n ( w / π π π w We have two square roots of w. Where is the other root? Same modulus but the argument of π radians is added because we have a square root which means we add π π radians. π 5π w + π π The two values of w are w and 5π w.. (a Using FOIL we have π
11 Complete Solutions to Examination Questions 0 ( + j( j j + j j + j+ 5+ j For the quotient we have to multiply the numerator and denominator by the complex conjugate of j which is + j. + j ( + j( + j + j+ j+ j j + + 5j 5 + j (b This is the same complex number as part (b of question 0. From that solution we have π π j The π π is not the principal argument but is the principal argument because the principal argument lies between π and π. π π (c Remember 5 has a modulus of 5 and an argument of π radians. 5 5 π The given quartic equation w + ( 5 0 can be rearranged to w 5 w ( 5 We have four roots of 5. Let w be the first root. By applying De Moivre s theorem we have
12 Complete Solutions to Examination Questions 0 w ( ( 5 5 ( π / n n 5 π Because ( r θ r ( nθ π π / ( 5 5 Because ( How do we find the other three roots? Add π π radians each time. Let and be the other three roots. w, w w π π π w π π 5π w π π 7π w All these roots lie on the circle with centre origin and radius 5 as illustrated below: Im w w Re - w w - -. (a The complex number i in polar form is π i + (, arg( i tan ( 5 or π π i Therefore i e. i (b The complex number e π in rectangular form is iπ e cos π isin π 0 ( (
13 Complete Solutions to Examination Questions 0 (c We first write the complex number + i in polar form: + i + / + ( / We have + i ( 5. Therefore arg i tan tan 5 60 ( ( 60 5 ( ( (d We can write 8 in polar form as 8 ( i ( The cube roots of 8 are given by / ( ( r / / The other roots are found by adding 60 0 : The cube roots of 8 are 0, r ( ( 0 r ( ( 0 r ( r ( and ( 0 r 0.
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