Express (5i)(-3i/5) in the form a+ib.
Hint. Multiply the coefficients and the i's separately, then substitute i^2=-1.
(5i)(-3i/5) = 5x(-3/5) x i^2 = -3 x (-1) = 3.
✦ Working through each part gives: 3 (i.e. 3+0i).
Express i^9 + i^19 in the form a+ib.
Hint. Reduce each exponent using the 4-cycle before adding.
i^9 = i^(4x2+1) = i. i^19 = i^(4x4+3) = i^3 = -i. Sum: i + (-i) = 0.
✦ Working through each part gives: 0.
Express i^(-39) in the form a+ib.
Hint. Add a multiple of 4 to the exponent to bring it into a familiar range, since i^4=1 does not change the value.
i^(-39) = i^(-39+40) = i^1 = i, adding 40 (a multiple of 4) to the exponent.
✦ Working through each part gives: i.
Express 3(7+i7) + i(7+i7) in the form a+ib.
Hint. Distribute both terms fully before combining real and imaginary parts.
3(7+7i) = 21+21i. i(7+7i) = 7i+7i^2 = 7i-7. Sum: (21-7) + (21+7)i = 14+28i.
✦ Working through each part gives: 14 + 28i.
Express (1-i) - (-1+i6) in the form a+ib.
Hint. Distribute the subtraction sign across both parts of the second complex number before combining.
Subtracting a complex number means flipping the sign of both its real and imaginary parts: (1-i) - (-1+6i) = (1-i) + (1-6i) = (1+1) + (-1-6)i = 2-7i.
✦ Working through each part gives: 2 - 7i.
Express (1/5 + i2/5) - (4 + i5/2) in the form a+ib.
Hint. Subtract real parts and imaginary parts separately, using a common denominator where needed.
Real part: 1/5-4 = 1/5-20/5 = -19/5. Imaginary part: 2/5-5/2 = 4/10-25/10 = -21/10.
✦ Working through each part gives: -19/5 - 21i/10.
Express [(1/3 + i7/3) + (4 + i1/3)] - (-4/3 + i) in the form a+ib.
Hint. Simplify the bracketed sum first, then subtract the last term.
Bracket: (1/3+4) + i(7/3+1/3) = 13/3 + i(8/3). Subtracting (-4/3+i): (13/3+4/3) + i(8/3-1) = 17/3 + i(5/3).
✦ Working through each part gives: 17/3 + 5i/3.
Express (1-i)^4 in the form a+ib.
Hint. Square (1-i) first to get a simple imaginary number, then square that result.
(1-i)^2 = 1-2i+i^2 = 1-2i-1 = -2i. So (1-i)^4 = (-2i)^2 = 4i^2 = -4.
✦ Working through each part gives: -4.
Express (1/3 + 3i)^3 in the form a+ib.
Hint. Expand using (a+b)^3 = a^3+3a^2b+3ab^2+b^3 with a=1/3, b=3i, then substitute powers of i.
a^3=1/27. 3a^2b = 3(1/9)(3i) = i. 3ab^2 = 3(1/3)(9i^2) = 3(1/3)(-9) = -9. b^3 = 27i^3 = -27i. Sum: (1/27-9) + (1-27)i = -242/27 - 26i.
✦ Working through each part gives: -242/27 - 26i.
Express (-2 - i/3)^3 in the form a+ib.
Hint. Factor out a negative sign to work with (2+i/3)^3 first, since cubing is an odd operation: (-x)^3=-x^3.
For (2+i/3)^3 with a=2,b=1/3: real part a^3-3ab^2 = 8-2/3 = 22/3; imaginary part 3a^2b-b^3 = 4-1/27 = 107/27. So (2+i/3)^3 = 22/3+(107/27)i, and (-2-i/3)^3 = -22/3-(107/27)i.
✦ Working through each part gives: -22/3 - (107/27)i.
Find the multiplicative inverse of 4 - 3i.
Hint. Use z^(-1) = z-bar/|z|^2.
z-bar = 4+3i. |z|^2 = 16+9 = 25. z^(-1) = (4+3i)/25.
✦ Working through each part gives: 4/25 + 3i/25.
Find the multiplicative inverse of √5 + 3i.
Hint. Use z^(-1) = z-bar/|z|^2.
z-bar = root5-3i. |z|^2 = 5+9 = 14. z^(-1) = (root5-3i)/14.
✦ Working through each part gives: root5/14 - 3i/14.
Find the multiplicative inverse of -i.
Hint. Use z^(-1) = z-bar/|z|^2, or simply check which number multiplies -i to give 1.
z=-i=0-i, so z-bar=0+i=i, and |z|^2=0+1=1. z^(-1) = i/1 = i. Check: (-i)(i) = -i^2 = 1, confirmed.
✦ Working through each part gives: i.
Express [(3+i√5)(3-i√5)] / [(√3+√2i) - (√3-i√2)] in the form a+ib.
Hint. The numerator is a product of conjugate-like factors (difference of squares); the denominator's two square-root terms cancel, leaving a purely imaginary denominator.
Numerator: (3+i-root5)(3-i-root5) = 9 - (i-root5)^2 = 9 - (i^2)(5) = 9-(-5) = 14. Denominator: (root3+root2 i) - (root3-i-root2) = root3+root2 i - root3 + i-root2 = 2-root2 i. So the expression is 14/(2-root2 i) = 7/(-root2 i) x (i/i) ... more directly, 1/i=-i, so 7/(root2 i) = (7/root2)(-i) = -7i/root2 = -(7root2/2)i.
✦ Working through each part gives: -(7√2/2)i (i.e. 0 - (7√2/2)i).
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.
Pair self-study with a tutor, a live course or a coaching centre.
