Evaluate: [i^18 + (1/i)^25]^3.
Hint. Reduce i^18 and (1/i)^25 separately using the 4-cycle before adding and cubing.
i^18 = i^(4x4+2) = i^2 = -1. i^25 = i^(4x6+1) = i, so (1/i)^25 = 1/i^25 = 1/i = -i. Sum inside brackets: -1+(-i) = -1-i. Cube: (-1-i)^3 = -(1+i)^3. (1+i)^2=2i, so (1+i)^3=(1+i)(2i)=2i+2i^2=-2+2i. So -(1+i)^3 = 2-2i.
✦ Working through each part gives: 2 - 2i.
For any two complex numbers z1 and z2, prove that Re(z1 z2) = Re z1 . Re z2 - Im z1 . Im z2.
Hint. Write z1=a+ib, z2=c+id, expand the product, and read off the real part.
z1z2 = (a+ib)(c+id) = (ac-bd) + i(ad+bc). By definition, Re(z1z2) = ac-bd, which is exactly (Re z1)(Re z2) - (Im z1)(Im z2) = ac-bd.
✦ Working through each part gives: re(z1z2) = Re z1 . Re z2 - Im z1 . Im z2, confirmed directly from the multiplication formula.
Reduce (1/(1-4i) - 2/(1+i))((3-4i)/(5+i)) to standard form.
Hint. Simplify each fraction separately by multiplying by its own conjugate first, then combine and multiply by the last factor.
1/(1-4i) = (1+4i)/17. 2/(1+i) = 1-i. Difference: (1+4i)/17-(1-i) = (1+4i-17+17i)/17 = (-16+21i)/17. Next, (3-4i)/(5+i) = (3-4i)(5-i)/26 = (11-23i)/26. Multiplying: (-16+21i)(11-23i) / (17x26) = (307+599i)/442, after expanding the numerator (-176+368i+231i+483) = 307+599i.
✦ Working through each part gives: (307 + 599i)/442.
If x - iy = √[(a-ib)/(c-id)], prove that (x²+y²)² = (a²+b²)/(c²+d²).
Hint. Square both sides first, then take the modulus of both sides rather than expanding directly — modulus of a quotient is the quotient of moduli.
Squaring: (x-iy)^2 = (a-ib)/(c-id). Taking modulus of both sides: |x-iy|^2 = |a-ib|/|c-id|, i.e. x^2+y^2 = root(a^2+b^2)/root(c^2+d^2). Squaring again: (x^2+y^2)^2 = (a^2+b^2)/(c^2+d^2).
✦ Working through each part gives: (x²+y²)² = (a²+b²)/(c²+d²), proved via modulus of a squared quotient.
If z1 = 2-i, z2 = 1+i, find |(z1+z2+1)/(z1-z2+1)|.
Hint. Compute the numerator and denominator as single complex numbers first, then simplify the ratio before taking its modulus.
z1+z2+1 = (2-i)+(1+i)+1 = 4. z1-z2+1 = (2-i)-(1+i)+1 = 2-2i. Ratio: 4/(2-2i) = 2/(1-i) = 2(1+i)/2 = 1+i. Modulus: |1+i| = root2.
✦ Working through each part gives: √2.
If a+ib = (x+i)²/(2x²+1), prove that a²+b² = (x²+1)²/(2x²+1)².
Hint. Take the modulus of both sides directly, using |z^2|=|z|^2 for the numerator.
|a+ib|^2 = |x+i|^4/(2x^2+1)^2, since |(x+i)^2| = |x+i|^2. |x+i|^2 = x^2+1. So a^2+b^2 = (x^2+1)^2/(2x^2+1)^2.
✦ Working through each part gives: a²+b² = (x²+1)²/(2x²+1)², proved via modulus of a squared complex number.
Let z1 = 2-i, z2 = -2+i. Find (i) Re(z1 z2 / z1-bar), (ii) Im(1/(z1 z1-bar)).
Hint. For (ii), remember z times its own conjugate is always a real number, so the fraction is automatically real before you even simplify it.
(i) z1z2 = (2-i)(-2+i) = -4+2i+2i-i^2 = -3+4i. z1-bar=2+i. Dividing: (-3+4i)/(2+i) = (-3+4i)(2-i)/5 = (-6+3i+8i+4)/5 = (-2+11i)/5, so Re = -2/5. (ii) z1z1-bar = |z1|^2 = 4+1=5, a real number, so 1/(z1z1-bar) = 1/5 is purely real, giving Im = 0.
✦ Working through each part gives: (i) Re = -2/5. (ii) Im = 0.
Find the real numbers x and y if (x - iy)(3 + 5i) is the conjugate of -6 - 24i.
Hint. The conjugate of -6-24i is -6+24i. Expand the left side and equate real and imaginary parts against it.
Conjugate of -6-24i is -6+24i. Expanding (x-iy)(3+5i) = (3x+5y) + i(5x-3y). Equating: 3x+5y=-6 and 5x-3y=24. Solving simultaneously (eliminate y): 9x+15y=-18 and 25x-15y=120; adding gives 34x=102, so x=3. Then 3(3)+5y=-6 gives y=-3.
✦ Working through each part gives: x = 3, y = -3.
Find the modulus of (1+i)/(1-i) - (1-i)/(1+i).
Hint. Simplify each fraction to a single power of i first — both reduce very cleanly.
(1+i)/(1-i) = (1+i)^2/2 = 2i/2 = i. (1-i)/(1+i) = (1-i)^2/2 = -2i/2 = -i. Difference: i-(-i) = 2i. Modulus: |2i| = 2.
✦ Working through each part gives: 2.
If (x+iy)³ = u+iv, then show that u/x + v/y = 4(x²-y²).
Hint. Expand (x+iy)^3 fully to identify u and v separately, then form each ratio.
(x+iy)^3 = (x^3-3xy^2) + i(3x^2y-y^3), so u=x^3-3xy^2 and v=3x^2y-y^3. u/x = x^2-3y^2. v/y = 3x^2-y^2. Sum: (x^2-3y^2)+(3x^2-y^2) = 4x^2-4y^2 = 4(x^2-y^2).
✦ Working through each part gives: u/x + v/y = 4(x²-y²), confirmed by dividing each part separately.
If α and β are different complex numbers with |β| = 1, find |(β-α)/(1-α-bar β)|.
Hint. Since |β|=1, β times its own conjugate is 1 — use this to show the numerator and denominator have equal modulus-squared, term by term.
Since |beta|=1, beta.beta-bar=1. |1-alpha-bar.beta|^2 = 1-alpha.beta-bar-alpha-bar.beta+|alpha|^2|beta|^2 = 1-alpha.beta-bar-alpha-bar.beta+|alpha|^2 (using |beta|^2=1). Separately |beta-alpha|^2 = |beta|^2-beta.alpha-bar-alpha.beta-bar+|alpha|^2 = 1-alpha-bar.beta-alpha.beta-bar+|alpha|^2. Both expressions are identical term by term, so the two moduli are equal.
✦ Working through each part gives: the value is 1 (the two moduli are always equal when |β|=1).
Find the number of non-zero integral solutions of the equation |1-i|^x = 2^x.
Hint. Compute |1-i| first as a power of 2, then compare exponents directly.
|1-i| = root2 = 2^(1/2). So (2^(1/2))^x = 2^x gives 2^(x/2) = 2^x, so x/2=x, meaning x=0. This is the only solution, and it is zero, not non-zero.
✦ Working through each part gives: 0 (there are no non-zero integral solutions).
If (a+ib)(c+id)(e+if)(g+ih) = A+iB, then show that (a²+b²)(c²+d²)(e²+f²)(g²+h²) = A²+B².
Hint. Take the modulus of both sides of the given equation, using the fact that modulus is multiplicative across any number of factors.
Taking modulus of both sides: |a+ib||c+id||e+if||g+ih| = |A+iB|. Squaring both sides: |a+ib|^2|c+id|^2|e+if|^2|g+ih|^2 = |A+iB|^2, i.e. (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2+B^2.
✦ Working through each part gives: (a²+b²)(c²+d²)(e²+f²)(g²+h²) = A²+B², proved via modulus multiplicativity.
If [(1+i)/(1-i)]^m = 1, then find the least positive integral value of m.
Hint. Simplify the base fraction to a single power of i first, then find the smallest m making that power equal 1.
(1+i)/(1-i) = (1+i)^2/2 = 2i/2 = i. So the equation becomes i^m=1, which holds exactly when m is a multiple of 4. The least positive such m is 4.
✦ Working through each part gives: m = 4.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh104.pdf) — one numbered exercise (4.1, 14 questions) plus the chapter's Miscellaneous Exercise (14 questions); the chapter has no Exercise 4.2/4.3, no discriminant/quadratic-formula-for-complex-roots content, and no polar-form content despite Section 4.5's title, matching CBSE's own 'formative-only' classification for both topics under this chapter. Several stacked-fraction questions in the Miscellaneous Exercise (Q1, Q3, Q4, Q7, Q9, Q11) were re-rendered directly from the PDF at 300dpi after raw text extraction garbled the nested fractions, most notably Q1 (confirmed as [i^18+(1/i)^25]^3, not the garbled '25/18i/1' the raw extraction suggested). Questions are referenced from the NCERT textbook for identification.
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