Question 1
Write (5i)(−3i/5) in the form a+ib.
Steps only
(5i)(−3i/5)=−3i²
i²=−1
Answer
3.
CLASS 11 · MATHEMATICS · CHAPTER 4
Complex Numbers and Quadratic Equations
Simplify complex numbers with powers of i, arithmetic, conjugates and modulus properties. Each answer begins with short working; open the explanation for the reasoning.
28 worked answersAll 1 exercise setsAll 14 end-of-chapter questions
What you will learn
- Use i²=−1 and the repeating cycle of powers of i.
- Add, subtract, multiply and divide complex numbers.
- Find multiplicative inverses using conjugates.
- Apply modulus properties to solve identities.
28 worked answers · short steps first, explanations when you need them
Exercise 4.1
Textbook pages 82–83
Question 2
Write i⁹+i¹⁹ in the form a+ib.
Steps only
i⁹=i⁸i=i
i¹⁹=i¹⁶i³=−i
i+(−i)=0
Answer
0.
Question 3
Simplify i⁻³⁹.
Steps only
−39=4(−10)+1
i⁻³⁹=i
Answer
i.
Question 4
Write 3(7+7i)+i(7+7i) in the form a+ib.
Steps only
3(7+7i)=21+21i
i(7+7i)=7i+7i²=−7+7i
Add real and imaginary parts
Answer
14+28i.
Question 5
Simplify (1−i)−(−1+6i).
Steps only
1−i+1−6i
(1+1)+(−i−6i)
Answer
2−7i.
Question 6
Write (1/5+2i/5)−(4+5i/2) in a+ib form.
Steps only
Real part: 1/5−4=−19/5
Imaginary part: 2/5−5/2=4/10−25/10=−21/10
Answer
−19/5−21i/10.
Question 7
Simplify [(1/3+7i/3)+(4+i/3)]−(−4/3+i).
Steps only
Inside bracket: 13/3+8i/3
Subtracting −4/3+i adds 4/3 and subtracts i
Real: 17/3; imaginary: 5/3
Answer
17/3+5i/3.
Question 8
Find (1−i)⁴.
Steps only
(1−i)²=1−2i+i²=−2i
(1−i)⁴=(−2i)²=4i²=−4
Answer
−4.
Question 9
Write (1/3+3i)³ in a+ib form.
Steps only
(a+ib)³=(a³−3ab²)+i(3a²b−b³)
a=1/3, b=3
Real=1/27−9=−242/27; imaginary=1−27=−26
Answer
−242/27−26i.
Question 10
Write (−2−i/3)³ in a+ib form.
Steps only
a=−2, b=−1/3
Real=a³−3ab²=−8+2/3=−22/3
Imaginary=3a²b−b³=−4+1/27=−107/27
Answer
−22/3−107i/27.
Question 11
Find the multiplicative inverse of 4−3i.
Steps only
1/(4−3i)=(4+3i)/[(4−3i)(4+3i)]
Denominator=4²+3²=25
Answer
(4+3i)/25.
Question 12
Find the multiplicative inverse of √5+3i.
Steps only
Multiply by conjugate √5−3i
Denominator=(√5)²+3²=14
Answer
(√5−3i)/14.
Question 13
Find the multiplicative inverse of −i.
Steps only
(−i)i=−i²=1
Answer
i.
Question 14
Write [(3+i√5)(3−i√5)]/[(√3+√2i)−(√3−i√2)] in a+ib form.
Steps only
Numerator=3²+5=14
Denominator=2√2 i
14/(2√2i)=7/(√2i)=−7i/√2
Answer
−7√2 i/2.
Miscellaneous Exercise on Chapter 4
Textbook pages 85–86
Question 1
Evaluate [i¹⁸+(1/i)²⁵]³.
Steps only
i¹⁸=−1
1/i=−i, so (1/i)²⁵=−i
(−1−i)³=2−2i
Answer
2−2i.
Question 2
Prove Re(z₁z₂)=Re(z₁)Re(z₂)−Im(z₁)Im(z₂).
Steps only
Let z₁=a+ib and z₂=c+id
z₁z₂=(ac−bd)+i(ad+bc)
Take the real part: ac−bd
Answer
The identity is proved.
Question 3
Reduce (1/(1−4i)−2/(1+i))((3−4i)/(5+i)) to a+ib form.
Steps only
1/(1−4i)=(1+4i)/17
2/(1+i)=1−i
(3−4i)/(5+i)=(11−23i)/26
Multiply (−16+21i)(11−23i)/(17·26)
Answer
307/442+599i/442.
Question 4
If x−iy=√[(a−ib)/(c−id)], prove (x²+y²)²=(a²+b²)/(c²+d²).
Steps only
Square both sides: (x−iy)²=(a−ib)/(c−id)
Take modulus squared
|x−iy|⁴=(a²+b²)/(c²+d²)
Answer
The identity is proved (for c²+d²≠0).
Question 5
For z₁=2−i and z₂=1+i, find |(z₁+z₂+1)/(z₁−z₂+1)|.
Steps only
Numerator=4
Denominator=2−2i
Quotient=4/(2−2i)=1+i
|1+i|=√2
Answer
√2.
Question 6
If a+ib=(x+i)²/(2x²+1), prove a²+b²=(x²+1)²/(2x²+1)².
Steps only
a²+b²=|a+ib|²
=|(x+i)²|²/(2x²+1)²
|x+i|²=x²+1
Answer
The identity is proved.
Question 7
Let z₁=2−i and z₂=−2+i. Find (i) Re(z₁z₂/conjugate(z₁)); (ii) Im(1/(z₁ conjugate(z₁))).
Steps only
z₁z₂=−3+4i and conjugate(z₁)=2+i
(−3+4i)/(2+i)=(−2+11i)/5
z₁conjugate(z₁)=|z₁|²=5
Answer
- (i) −2/5;
- (ii) 0.
Question 8
Find real x,y if (x−iy)(3+5i) is the conjugate of −6−24i.
Steps only
Conjugate is −6+24i
Product=(3x+5y)+i(5x−3y)
3x+5y=−6; 5x−3y=24
Solve to get x=3, y=−3
Answer
x=3, y=−3.
Question 9
Find |(1+i)/(1−i)−(1−i)/(1+i)|.
Steps only
(1+i)/(1−i)=i
(1−i)/(1+i)=−i
|i−(−i)|=|2i|=2
Answer
2.
Question 10
If (x+iy)³=u+iv, show u/x+v/y=4(x²−y²), where x and y are non-zero.
Steps only
u=x³−3xy²
v=3x²y−y³
u/x+v/y=x²−3y²+3x²−y²
Answer
The identity is proved.
Question 11
If α≠β and |β|=1, find |(β−α)/(1−conjugate(α)β)|.
Steps only
conjugate(β)=1/β because |β|=1
1−conjugate(α)β=β(conjugate(β)−conjugate(α))
Denominator modulus=|β−α|
Answer
1.
Question 12
Find the number of non-zero integer solutions of |1−i|ˣ=2ˣ.
Steps only
|1−i|=√2
2ˣ⁄²=2ˣ implies x/2=x
Only x=0, which is excluded
Answer
0.
Question 13
If (a+ib)(c+id)(e+if)(g+ih)=A+iB, prove (a²+b²)(c²+d²)(e²+f²)(g²+h²)=A²+B².
Steps only
Take modulus squared of both sides
|zw|=|z||w|
|a+ib|²=a²+b²; |A+iB|²=A²+B²
Answer
The identity is proved.
Question 14
If ((1+i)/(1−i))ᵐ=1, find the least positive integer m.
Steps only
(1+i)/(1−i)=i
iᵐ=1
i⁴=1 and i,i²,i³ are not 1
Answer
m=4.