H2 Mathematics: Complex Numbers 1
2010 Meridian Junior College
H2 Mathematics (9740) Complex Numbers 1 Tutorial: Complex Numbers in Cartesian Form 1
Find the complex number z such that Solution: z−2 = 1+ i z z − 2 = (1 + i )z z − (1 + i )z = 2 z (1 − 1 − i ) = 2 2 = 2i z= −i
2
z−2 = 1 + i. z
Key Learning Points: • Solving simple equations • Simply make z the subject • Use GC to check that answer obtained is correct • Avoid letting z = x + iy as this will complicate matters
The complex number z and w are such that w = 1 + a i, z = – b + i , where a and b are real and positive. Given that wz* = 3 – 4i, find the exact values of a and b. Solution: wz* = 3 – 4i (1 + ai )(− b − i ) = 3 − 4i − b − i − abi + a = 3 − 4i −b + a + ( −1 − ab ) i = 3 − 4i Equating real parts: − b + a = 3 a = 3 + b ---------- (1) Equating imaginary parts: − 1 − ab = −4 ab = 3 ---------- (2) Substitute (1) into (2): Key Learning Points: (3 + b)b = 3 • Equality of complex numbers [two complex numbers are equal ⇔ their b 2 + 3b − 3 = 0 real and imaginary parts are equal − 3 ± 9 − 4(1)(−3) • Equate real and imaginary parts to b= obtain pair of equations 2 • Solve pair of equations simult to obtain − 3 ± 21 b= a and b 2 • Students have to choose the correct value and give the relevant reason why − 3 + 21 Since b > 0, ∴ b = [in this case since b > 0] 2 − 3 + 21 3 + 21 From equation (1): a = 3 + = 2 2
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H2 Mathematics: Complex Numbers 1
3
2010 Meridian Junior College
Given that z = w + 3i + 2 and z 2 − wi + 5 − 2i = 0 , find the complex numbers z and w in the form a + bi, where a and b are real numbers. Solution: z = w + 3i +2 ---------- (1) z2 – wi + 5 – 2i = 0 ---------- (2) From (1): w = z – 3i – 2 ---------- (3) Substitute (3) into (2): z 2 − ( z − 3i − 2 )i + 5 − 2i = 0 z 2 − zi − 3 + 2i + 5 − 2i = 0
Key Learning Points: • Solving simultaneous equations • Use the substitution method [recommended] • Sub linear into quadratic— careful when multiplying out • Give solutions as pairs—do not give solutions as a list [say:
∴ z = 2i , w = – 2 – i, z = −i , w = – 2 – 4i] this gives the
z − zi + 2 = 0 2
i ± (i) 2 − 4(1)(2) i ± − 9 i ± 3i z= = = 2 2 2 ∴ z = 2i or z = −i
From equation (3): when z = 2i,
•
impression that there are 4 possible pairs when in actual fact there are only two!! A good point to illustrate the idea that polynomial equations with non-real coefficients give rise to roots which may not be conjugate pairs
w = 2i – 3i – 2 = – 2 – i
when z = –i, w = –i – 3i – 2 = – 2 – 4i ∴ z = 2i , w = – 2 – i z = −i , w = – 2 – 4i
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H2 Mathematics: Complex Numbers 1
4
2010 Meridian Junior College
Given that ( x + yi) 2 = −5 + 12i , where x and y are real, find the possible values of x + yi. Hence, solve z 2 + 4 z = −9 + 12i . Solution: ( x + yi)2 = −5 + 12i
[Without GC]: x 2 − y 2 + 2 xyi = −5 + 12i
x 2 − y 2 = −5 4
& 2 xy = 12 [comparing real and imaginary parts]
2
x + 5 x − 36 = 0 x2 + 9 x2 − 4 = 0
(
)(
)
x = ±2 [since x is real] For x = 2, y = 3 . For x = −2, y = −3 . ∴ the solutions are: 2 + 3i or − 2 − 3i .
[OR Using GC]: x + yi = ± − 5 + 12i = ± (2 + 3i ) z 2 + 4 z = −9 + 12i (z + 2 )2 − 4 = −9 + 12i
(z + 2 )2 = −5 + 12i (z + 2 ) = ± − 5 + 12i z + 2 = (2 + 3i )
∴ z = 3i
or or
z + 2 = −2 − 3i z = −4 − 3i
Second Part • Hence means we use previous part • Trick is to complete the square—then the use of previous part becomes obvious • Students should remember the
Key Learning Points: • Can be solved directly using GC [GC will only give the positive root— remember to also give the negative root] • Algebraically: This is about equality of complex numbers [Same idea as Q2 • Equality of complex numbers [two complex numbers are equal ⇔ their real and imaginary parts are equal • Equate real and imaginary parts to obtain pair of equations • Solve pair of equations simult to obtain x and y • • •
2
Why is x + 9 = 0 rejected? [since x ∈ ] Solutions are to be given in the form x + yi Use GC to check
± blahblah • •
IF question had said [Hence or otherwise] then students could use the quadratic formula to solve directly. Use GC to check
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H2 Mathematics: Complex Numbers 1
5
2010 Meridian Junior College
2006 MJC Prelim/I/2 5 7 − i as two of its roots, in the form 4 4 az 3 + bz 2 + cz + d = 0 where a, b, c and d are real coefficients to be determined.
Obtain a cubic equation having −2 and
Solution: 5 7 5 7 − i is one of the roots + i is also a root [since the polynomial has 4 4 4 4 real coefficients—insist that the students give this reason]
Since
5 7 A possible cubic equation is ( z + 2 ) z − + i 4 4
( z + 2) ( 2z 2 − 5z + 4) = 0
5 7 i =0 z− − 4 4
2z 3 − z 2 − 6z + 8 = 0 ∴ a = 2, b = −1, c = −6, d = 8
Note: Any equation of the form k ( 2 z 3 − z 2 − 6 z + 8 ) = 0 where k ∈
| {0} is acceptable.
Key Learning Points: • Solving polynomial equations with real coefficients • Make clear to students that complex roots appear as conjugate pairs for polynomial equations with real coefficients [but this may not be the case for those with non-real coefficients] •
Students have to explain their choice of the roots
5 7 5 7 + i [Since − i is one of 4 4 4 4
5 7 + i is also a root [since the polynomial has real 4 4
coefficients—insist that the students give this reason]] •
Ask the students if 2 z − z − 6 z + 8 = 0 is the only possible solution; then talk about the fact that infinitely many solutions exist which are of the form: 3
2
k ( 2 z 3 − z 2 − 6 z + 8 ) = 0 where k ∈
•
| {0}
Use GC to check
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H2 Mathematics: Complex Numbers 1
6
2010 Meridian Junior College
[A graphic calculator is not to be used in answering this question.] 2 5 One of the roots of the equation 9 x 3 + 6 x 2 + 5 x + 58 = 0 is − i . Find the other roots of 3 3 the equation. Solution: Method 1
2 5 2 5 − i is one of the roots + i is also a root [since the polynomial has 3 3 3 3 real coefficients—insist that the students give this reason]
Since
Hence, x −
2 5 − i 3 3
x−
2 5 + i 3 3
=
= x2 −
4 29 is a factor of x+ 3 9
9 x 3 + 6 x 2 + 5 x + 58 .
4 29 x+ 3 9 By comparing coefficients of x3 : a = 9 9 x 3 + 6 x 2 + 5 x + 58 = x 2 −
( ax + b )
x 0 : b = 18 ∴ 9 x 3 + 6 x 2 + 5 x + 58 = x 2 −
4 29 x+ 3 9
( 9 x + 18)
∴ The other roots of the equation are x = -2 or
Key Learning Points: • Read rubrics of question [No GC allowed] • Similar to Q5—about solving cubic equations with real coefficients— same learning points • Reason for choosing the conjugate as a root has to be given • Use GC to check
2 5 + i. 3 3
Method 2 Let f(x) = 9 x 3 + 6 x 2 + 5 x + 58 By trial and error, f(–2) = 0
Since
2 5 − i is one of the roots 3 3
x + 2 is a factor of f(x) x = −2 is a root of f(x) = 0 2 5 + i is also a root. 3 3
∴ The other roots of the equation are x = –2 and x =
2 5 + i. 3 3
[Note: If the use of GC were allowed, then we can use the APPS [Poly Root Finder] to obtain the roots easily.]
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H2 Mathematics: Complex Numbers 1
7
2010 Meridian Junior College
N2005/II/1 [A graphic calculator is not to be used in answering this question.]
Verify that z = i is a root of the equation z 4 − 2 z 3 + 6 z 2 − 2 z + 5 = 0 . Hence determine the other roots.
Solution: Let f ( z ) = z 4 − 2 z 3 + 6 z 2 − 2 z + 5 = 0
Note: f ( i ) = i4 – 2(i)3 + 6(i)2 – 2i + 5 = 1 + 2i – 6 – 2i + 5 [This line has to be shown] =0 ∴ z = i is a root of the equation. (verified) Since z = i is a root, z = −i is also a root [since the polynomial has real coefficients—insist that the students give this reason] ∴ f ( z ) = z4 – 2z3 + 6z2 – 2z + 5 = (z2 + 1) (z2 + az +b)
By comparing coefficient of z3: –2 = a By comparing constant: 5 = b ∴ f ( z ) = z4 – 2z3 + 6z2 – 2z + 5 = (z2 + 1) (z2 –2z + 5) = 0 z2 –2z + 5 = 0 2 ± 4 − 4(1)(5) z= 2 2 ± 4i = 2 = 1 ± 2i
For:
∴ The other roots of the equation are z = – i or z = 1 + 2i or z = 1 − 2i . Key Learning Points: • Similar to Q6 –No GC allowed • This questions is also about roots of polynomials with real coefficients; reason for choosing the conjugate as a root has to be given • Verifying means writing down enough working to convince the examiner— students cannot just say: From GC • •
f ( i ) = 0;they need to write the line 1 +
2i – 6 – 2i + 5 = 0,thus …. Again students should use GC to check their answers IF GC was allowed then solutions could be found directly using the GC
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H2 Mathematics: Complex Numbers 1
8
2010 Meridian Junior College
Find the real value of k such that 2 x 4 − 4 x 3 + 3 x 2 + 2 x + k = 0 has a complex root 1 − i . Hence, factorise 2 x 4 − 4 x 3 + 3 x 2 + 2 x + k completely. Solution: Let f(x) = 2 x 4 − 4 x 3 + 3 x 2 + 2 x + k 4 3 2 Note: f(1– i) = 0 2 (1 − i ) − 4 (1 − i ) + 3 (1 − i ) + 2 (1 − i ) + k = 0
2(− 4) − 4(− 2 − 2i ) + 3(− 2i ) + 2 − 2i + k = 0 k = −2 Since x = 1 − i is a root of the equation, x = 1 + i is also a root [since the polynomial has real coefficients—insist that the students give this reason]
(
f(x) = 2 x 4 − 4 x 3 + 3 x 2 + 2 x − 2 = (x − 1 + i )( x − 1 − i ) 2 x 2 + ax + b
(
)(
)
)
= x 2 − 2 x + 2 2 x 2 + ax − 1
By comparing coefficient of x3: – 4 = a – 4
( = (x
)( ) − 2 x + 2 )( 2 x + 1)(
∴ f(x) = x 2 − 2 x + 2 2 x 2 − 1
9
2
a=0
)
Similar to Q5-7— same concepts and learning points
2x − 1
Given that x = 2 + ai is a root of the equation x 3 − x 2 + ( a 2 − 8 ) x + 12 + 3a 2 = 0 , where
a∈
, find the other two roots.
Solution: Since x = 2 + ai is a root of the equation, x = 2 − ai is also a root [since the polynomial has real coefficients—insist that the students give this reason] . Hence, x − ( 2 + ai ) x − ( 2 − ai ) = = x 2 − 4 x + 4 + a 2 is a quadratic factor of x3 − x 2 + ( a 2 − 8 ) x + 12 + 3a 2 .
∴ x3 − x 2 + ( a 2 − 8 ) x + 12 + 3a 2 = ( x 2 − 4 x + 4 + a 2 ) ( x + k )
By comparing constant term, k = 3 . ∴ The other two roots are x = 2 − ai or x = −3 . Similar to Q5-8—Same concepts and learning points
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H2 Mathematics: Complex Numbers 1
2010 Meridian Junior College
10 N2007/I/3(b) The complex number w is such that ww * +2 w = 3 + 4i, where w * is the complex conjugate of w . Find w in the form a + ib, where a and b are real. Solution: ww * +2 w = 3 + 4i, Letting w = a + ib we get: a 2 + b 2 + 2 ( a + bi ) = 3 + 4i a 2 + b 2 + 2a + 2bi = 3 + 4i ----------(1) Comparing real and imaginary parts of (1) we get: a 2 + b 2 + 2a = 3 --------------------(2) 2b = 4 -------------------------------(3) From (3): b = 2 Substituting b = 2 we get:
a 2 + 2a + 4 = 3
a 2 + 2a + 1 = 0
( a + 1)2 = 0
a = −1
∴ w = −1 + 2i Key Learning Points: • Equality of complex numbers • Equality of complex numbers [two complex numbers are equal ⇔ their real and imaginary parts are equal • Let w = a + ib to simply your work w = a + ib • Equate real and imaginary parts to obtain pair of equations • Solve pair of equations simultaneously to obtain a and b • Students should remember to give w in the form w = a + ib
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H2 Mathematics: Complex Numbers 1
2010 Meridian Junior College
11 It is given that z1 = 3 − 2i is a root of the equation 2 z 3 + az 2 + bz + 39 = 0 , where a and b are real constants. (i) Show that ( z − z1 )( z − z2 ) = z 2 − 6 z + 13 where z2 is another non-real root of the equation. (ii) In either order, find: (a) the value of a and of b, (b) the third root, z3 , of the equation. Solution: (i) Note: If z1 = 3 − 2i is a root then z2 = 3 + 2i is also a root of this equation [since the polynomial has real coefficients—insist that the students give this reason] Thus: ( z − z1 )( z − z2 ) = ( z − ( 3 − 2i ) ) ( z − ( 3 + 2i ) )
= z 2 − 3z − 2iz − 3z + 2iz + ( 9 + 4 )
= z 2 − 6 z + 13 [Shown] (ii) Note: 2 z 3 + az 2 + bz + 39 = ( z 2 − 6 z + 13) ( 2 z + 3) [By inspection]
Thus: (a) a = −9 & b = 8 3 (b) z3 = − 2
= 2 z 3 −9 z 2 + 8 z + 39 Similar to Q5-8—Same concepts and learning points
___________________________________________________________________________ Assignment: 1
Find, in the form x + yi where x and y are real, the two complex numbers z z 3 4 satisfying the equations = + i and zz* = 5 . z* 5 5 Solution: Multiply the two equations: z 3 4 × zz* = + i × 5 z* 5 5 ∴ Using GC, z = ±(2 + i ) .
z 2 = 3 + 4i
z = ± 3 + 4i
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H2 Mathematics: Complex Numbers 1
2
2010 Meridian Junior College
N2004/I/12(a) Express (3 – i)2 in the form a + ib. Hence, or otherwise, find the roots of the equation (z + i)2 = –8 + 6i. Solution: Using GC, (3 – i)2 = 8 – 6i (z + i)2 = –8 + 6i (z + i)2 = – (8 – 6i) (z + i)2 = – (3 – i)2 (z + i)2 = i2 (3 – i)2 (z + i) = ± i (3 − i) Thus: ( z + i) = 1 + 3i or ( z + i) = −1 − 3i ∴ z = 1 + 2i
3
or
z = –1 – 4i
N2006/I/6 [A graphic calculator is not to be used in answering this question.]
Show that the equation z 4 − 2 z 3 + 6 z 2 − 8 z + 8 = 0 has a root of the form ki , where k is real. Hence solve the equation z 4 − 2 z 3 + 6 z 2 − 8 z + 8 = 0 . Solution: Let f ( z ) = z 4 − 2 z 3 + 6 z 2 − 8 z + 8 = 0 f(2i) = (2i)4 – 2(2i)3 + 6(2i)2 – 8(2i) + 8 = 0 Thus 2i is a root of this equation. (shown) Since 2i is a root, – 2i is also a root [since the polynomial has real coefficients—insist that the students give this reason]. ∴ f ( z ) = z 4 − 2 z 3 + 6 z 2 − 8 z + 8 = ( z 2 + 4 )( z 2 + az + b )
By comparing coefficient of z3: –2 = a By comparing constant: 8 = 4b
b=2
∴ f ( z ) = ( z 2 + 4 )( z 2 − 2 z + 2 ) = 0
For: z 2 − 2 z + 2 = 0 . z=
2 ± 4 − 4(1)(2) 2 ± 2i = = 1± i 2 2
∴ z = 2i or − 2i or 1 + i or 1 − i
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