Miscellaneous Exercise answers: Complex Numbers and Quadratic Equations

Open NCERT p. 85Matches NCERT’s answer

1. 18 = 4×4+2, so i¹⁸ = (i⁴)⁴ × i² = 1×(−1) = −1.
2. 1/i = −i, so (1/i)²⁵ = (−i)²⁵ = −i²⁵.
3. 25 = 4×6+1, so i²⁵ = i, hence (1/i)²⁵ = −i.
4. Add inside the bracket: −1 + (−i) = −1 − i.
5. Cube it: (−1−i)² = 1+2i+i² = 2i. Then (2i)(−1−i) = −2i−2i² = 2−2i.

Answer2 − 2i

Watch this explained “Two harder ones”, 9:23 into Why the powers of the new symbol run round a cycle of four

Open NCERT p. 85One way to think about it

1. Prove: Re (z₁z₂) = Re z₁ Re z₂ − Im z₁ Im z₂.
2. Let z₁ = a+ib and z₂ = c+id, so Re z₁=a, Im z₁=b, Re z₂=c, Im z₂=d.
3. Multiply: z₁z₂ = (a+ib)(c+id) = ac + iad + ibc + i²bd.
4. Since i² = −1, this is (ac−bd) + i(ad+bc).
5. The real part of z₁z₂ is ac−bd, which is exactly Re z₁ Re z₂ − Im z₁ Im z₂.

In shortProved: Re(z₁z₂) = ac − bd = Re z₁ Re z₂ − Im z₁ Im z₂.

Watch this explained “The one substitution”, 2:21 into Multiplying, then inverting: how division becomes possible

Open NCERT p. 86Matches NCERT’s answer

1. Find 1/(1−4i) by multiplying top and bottom by (1+4i): (1+4i)/17.
2. Find 2/(1+i) by multiplying top and bottom by (1−i): 2(1−i)/2 = 1−i.
3. Subtract: (1+4i)/17 − (1−i) = (1+4i−17+17i)/17 = (−16+21i)/17.
4. Multiply by (3−4i): (−16+21i)(3−4i) = −48+64i+63i+84 = 36+127i, so we now have (36+127i)/17.
5. Divide by (5+i): multiply top and bottom by (5−i). Numerator (36+127i)(5−i) = 307+599i. Denominator 17×26 = 442.

Answer307/442 + 599/442 i

Watch this explained “Dividing is not a new rule”, 11:40 into Multiplying, then inverting: how division becomes possible

Open NCERT p. 86One way to think about it

1. Given: x − iy = √((a − ib)/(c − id)). Prove: (x² + y²)² = (a² + b²)/(c² + d²).
2. Square both sides: (x−iy)² = (a−ib)/(c−id).
3. Take the conjugate of both sides too: (x+iy)² = (a+ib)/(c+id), since conjugating flips every i.
4. Multiply the two equations: (x−iy)²(x+iy)² = [(a−ib)(a+ib)] / [(c−id)(c+id)].
5. The left side is [(x−iy)(x+iy)]² = (x²+y²)², since (x−iy)(x+iy) = x²+y².
6. The right side is (a²+b²)/(c²+d²), since (a−ib)(a+ib)=a²+b² and (c−id)(c+id)=c²+d².
7. So (x²+y²)² = (a²+b²)/(c²+d²).

In shortProved.

Watch this explained “Five results, rarely proved”, 9:08 into Size and reflection: two quantities that turn algebra into geometry

Open NCERT p. 86Matches NCERT’s answer

1. z₁+z₂+1 = (2−i)+(1+i)+1 = 4.
2. z₁−z₂+1 = (2−i)−(1+i)+1 = 2−2i.
3. |4| = 4.
4. |2−2i| = √(2²+2²) = √8 = 2√2.
5. Divide: 4/(2√2) = 2/√2 = √2.

Answer√2

Watch this explained “Five results, rarely proved”, 9:08 into Size and reflection: two quantities that turn algebra into geometry

Open NCERT p. 86One way to think about it

1. Take the modulus of both sides: |a+ib| = |x+i|² / |2x²+1|, using that the size of a quotient is the quotient of the sizes.
2. |x+i| = √(x²+1), so |x+i|² = x²+1.
3. 2x²+1 is a positive real number, so |2x²+1| = 2x²+1.
4. So |a+ib| = (x²+1)/(2x²+1).
5. Square both sides: |a+ib|² = a²+b², so a²+b² = (x²+1)²/(2x²+1)².

In shortProved.

Watch this explained “Five results, rarely proved”, 9:08 into Size and reflection: two quantities that turn algebra into geometry

Open NCERT p. 86Matches NCERT’s answer

(i) Re(z1z2/z̄1)

1. z₁z₂ = (2−i)(−2+i) = −4+2i+2i−i² = −3+4i.
2. The conjugate of z₁ is 2+i.
3. Divide (−3+4i)/(2+i) by multiplying top and bottom by (2−i): numerator (−3+4i)(2−i) = −2+11i; denominator 4+1=5.
4. So the quotient is (−2+11i)/5, and its real part is −2/5.

Answer−2/5

(ii) Im(1/(z1z̄1))

1. z₁ times its own conjugate: (2−i)(2+i) = 4+1 = 5, an ordinary real number.
2. 1/5 is also a real number, so its imaginary part is 0.

Answer0

Watch this explained “What the identity saves”, 14:21 into Size and reflection: two quantities that turn algebra into geometry

Open NCERT p. 86Matches NCERT’s answer

1. The conjugate of −6 − 24i is −6 + 24i (change the sign of the i-part).
2. Expand the left side: (x − iy)(3 + 5i) = 3x + 5xi − 3yi − 5yi² = (3x + 5y) + i(5x − 3y), since i² = −1.
3. Match the real parts: 3x + 5y = −6.
4. Match the i-parts: 5x − 3y = 24.
5. Multiply the first equation by 3 and the second by 5: 9x + 15y = −18 and 25x − 15y = 120.
6. Add them: 34x = 102, so x = 3.
7. Put x = 3 into 3x + 5y = −6: 9 + 5y = −6, so 5y = −15 and y = −3.
8. Check: (3 + 3i)(3 + 5i) = 9 + 15i + 9i + 15i² = −6 + 24i. ✓

Answerx = 3, y = −3

Watch this explained “The equation, worked through”, 9:44 into Splitting a number into two parts, and when two such numbers agree

Open NCERT p. 86Matches NCERT’s answer

1. (1+i)/(1−i): multiply top and bottom by (1+i): (1+i)²/2 = (1+2i−1)/2 = 2i/2 = i.
2. (1−i)/(1+i): multiply top and bottom by (1−i): (1−i)²/2 = (1−2i−1)/2 = −2i/2 = −i.
3. Subtract: i − (−i) = 2i.
4. The modulus of 2i is √(0²+2²) = 2.

Answer2

Watch this explained “Clearing it by hand”, 9:13 into Multiplying, then inverting: how division becomes possible

Open NCERT p. 86One way to think about it

1. Show: u/x + v/y = 4(x² − y²).
2. Expand (x+iy)³ = x³ + 3x²(iy) + 3x(iy)² + (iy)³.
3. = x³ + 3ix²y − 3xy² − iy³ (using i²=−1 and i³=−i).
4. Group the plain and i-parts: (x³−3xy²) + i(3x²y−y³).
5. So u = x³−3xy² and v = 3x²y−y³.
6. u/x = x²−3y² and v/y = 3x²−y² (dividing every term by x, then by y).
7. Add: u/x+v/y = (x²−3y²)+(3x²−y²) = 4x²−4y² = 4(x²−y²).

In shortProved.

Watch this explained “Cubes, and a fourth power”, 10:53 into Which school algebra identities survive the enlargement, and why

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1. For any complex number z, z·z̄ = |z|². Since |β| = 1, ββ̄ = 1.
2. Top: |β − α|² = (β − α)(β̄ − ᾱ) = ββ̄ − ᾱβ − αβ̄ + αᾱ = 1 − ᾱβ − αβ̄ + |α|².
3. Bottom: |1 − ᾱβ|² = (1 − ᾱβ)(1 − αβ̄) = 1 − αβ̄ − ᾱβ + αᾱββ̄ = 1 − ᾱβ − αβ̄ + |α|², using ββ̄ = 1 again.
4. The two are equal, so |β − α| = |1 − ᾱβ|.
5. The bottom is not zero: if 1 − ᾱβ = 0 then |β − α| = 0, so α = β — but α and β are different.
6. So |(β − α)/(1 − ᾱβ)| = |β − α| ÷ |1 − ᾱβ| = 1.

Answer1

Watch this explained “Two quantities, one job”, 0:00 into Size and reflection: two quantities that turn algebra into geometry

Open NCERT p. 86Matches NCERT’s answer

1. Given: |1 − i|ˣ = 2ˣ.
2. |1−i| = √(1²+(−1)²) = √2.
3. So the equation is (√2)ˣ = 2ˣ, i.e. 2^(x/2) = 2ˣ.
4. 2 raised to two different powers can only be equal when the powers themselves are equal, so x/2 = x.
5. This gives x = 0, the only solution.
6. x = 0 is not a non-zero value, so there are no non-zero integral solutions.

Answer0

Watch this explained “The sum of two squares”, 1:00 into Size and reflection: two quantities that turn algebra into geometry

Open NCERT p. 86One way to think about it

1. The modulus of a product equals the product of the moduli: |z₁z₂z₃z₄| = |z₁||z₂||z₃||z₄|.
2. Here z₁=a+ib, z₂=c+id, z₃=e+if, z₄=g+ih, and their product is A+iB.
3. So |A+iB| = √(a²+b²) × √(c²+d²) × √(e²+f²) × √(g²+h²).
4. Square both sides: A²+B² = (a²+b²)(c²+d²)(e²+f²)(g²+h²).

In shortProved.

Watch this explained “What the identity saves”, 14:21 into Size and reflection: two quantities that turn algebra into geometry

Open NCERT p. 86Matches NCERT’s answer

1. Given: ((1 + i)/(1 − i))ᵐ = 1.
2. Simplify (1+i)/(1−i) by multiplying top and bottom by (1+i): (1+i)²/2 = 2i/2 = i.
3. So the equation becomes iᵐ = 1.
4. The powers of i repeat every 4 steps: i¹=i, i²=−1, i³=−i, i⁴=1.
5. The smallest positive m for which iᵐ=1 is m=4.

Answer4

Watch this explained “Two harder ones”, 9:23 into Why the powers of the new symbol run round a cycle of four