Q.(−i)(2i)(−81i)3=
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Complex Number Arithmetic
Complex Number Arithmetic: A First Look
Imagine you're trying to solve x2+1=0. You know that no real number squared gives −1. The square of any real number is either zero or positive. So this equation has no real solution. But what if we invent a number whose square is −1? That's exactly what mathematicians did — and that invention is the imaginary unit i, defined by:
i2=−1
A complex number is any number of the form a+bi, where a and b are real numbers. Here a is called the real part, and b is called the imaginary part. For example, 3+4i has real part 3 and imaginary part 4.
The name "imaginary" is unfortunate — these numbers are just as real (in the mathematical sense) as the numbers you already know. They're simply a different kind of number.
Why Bother?
Complex numbers let you solve equations that real numbers can't. Every polynomial equation — no matter how complicated — has a solution in the complex numbers. This is the Fundamental Theorem of Algebra, and it's one of the most important results in mathematics.
Arithmetic Operations
The rules are straightforward: treat i like a variable, but remember that i2=−1.
Addition and Subtraction
Add (or subtract) real parts with real parts, imaginary parts with imaginary parts.
(a+bi)+(c+di)=(a+c)+(b+d)i
(a+bi)−(c+di)=(a−c)+(b−d)i
Example: (2+3i)+(4−5i)=(2+4)+(3−5)i=6−2i
Multiplication
Multiply like binomials, then replace i2 with −1.
(a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+(ad+bc)i
Example: (2+3i)(4−5i)=2(4)+2(−5i)+3i(4)+3i(−5i)
=8−10i+12i−15i2
=8+2i−15(−1)
=8+2i+15=23+2i
The most common mistake: forgetting that i2=−1, not 1. Always check your final step.
Division
Division is trickier. The key idea: multiply numerator and denominator by the complex conjugate of the denominator.
The complex conjugate of a+bi is a−bi. When you multiply a complex number by its conjugate, you get a real number:
(a+bi)(a−bi)=a2−(bi)2=a2−b2i2=a2+b2
So to divide:
c+dia+bi=(c+di)(c−di)(a+bi)(c−di)=c2+d2(a+bi)(c−di)
Example: 4−5i2+3i=(4−5i)(4+5i)(2+3i)(4+5i)=16+258+10i+12i+15i2=418+22i−15=41−7+22i=−417+4122i
To divide quickly: multiply top and bottom by the conjugate of the denominator, then simplify. The denominator always becomes c2+d2, a positive real number.
The Big Picture …
Simplify step by step: first multiply the two simple factors, then cube the fraction, then multipl …
(−i)(2i)(−81i)3=256i.
Step 1: (−i)(2i)=−2i2=−2(−1)=2 (using i2=−1).
Step 2: (−81i)3=(−81)3i3=−5121×(−i)=512i (using i3=−i).
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Showing the 12 most recent of 94 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The multiplicative inverse of 5+3i is(a) 5−3i(b) 145+i⋅143(c) 145−i⋅143(d) 3−5i
›Reveal solutionSolution
The multiplicative inverse of a+ib is a2+b2a−ib; here a=5,b=3.
The multiplicative inverse of z=5+3i is z1. Rationalise by multiplying by the conjugate 5−3i: …
- CBSE 2026Set ANNUAL1 markMCQQ.The modulus of 1−i1+i−1+i1−i is(a) 2(b) -2(c) 1(d) -1
›Reveal solutionSolution
Simplify each fraction using the conjugate, subtract, then take the modulus of the resulting purely imaginary number.
1−i1+i=(1−i)(1+i)(1+i)2=1−i21+2i+i2=1+11+2i−1=22i=i …
- CBSE 2026Set ANNUAL1 markMCQQ.If (1−i1+i)m=1 then the least integral value of m is(a) 1(b) 2(c) 3(d) 4
›Reveal solutionSolution
Simplify the base to i, then find the smallest positive integer power of i equal to 1 (the order of i is 4).
As shown above, 1−i1+i=i. So the equation becomes im=1.
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- CBSE 2026Set ANNUAL1 markMCQQ.The value of i28 is —(a) i(b) −i(c) 1(d) −1
›Reveal solutionSolution
i28=1, option (c).
Recall i2=−1, so i4=(i2)2=(−1)2=1. Powers of i cycle with period 4: i,−1,−i,1,i,−1,…
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- CBSE 2026Set ANNUAL1 markQ.The conjugate of the complex number 2−3i is ______.
›Reveal solutionSolution
The conjugate of 2−3i is 2+3i.
For a complex number z=a+ib, the conjugate zˉ=a−ib — the sign of the imaginary part is reversed while the real part stays unchanged.
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- CBSE 2026Set ANNUAL1 markMCQQ.Multiplicative inverse of 1 + √3i is:(a) 1 - √3i(b) (1 - √3i)/2(c) (1 + √3i)/4(d) None of these
›Reveal solutionSolution
z⁻¹ = conjugate(z)/|z|²; for z = 1+√3i this gives (1−√3i)/4, which isn't among options (a)-(c) as written, so the correct choice is (d).
For a complex number z=1+3i, the multiplicative inverse is:
z−1=∣z∣2zˉ
Here zˉ=1−3i, and ∣z∣2=12+(3)2=1+3=4.
So:
z−1=41−3i
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- CBSE 2026Set ANNUAL1 markQ.The modulus of (1 + i)/(1 - i) is ..............
›Reveal solutionSolution
Rationalize (1+i)/(1-i) to a+bi form, or use |z1/z2| = |z1|/|z2| — both give modulus 1.
Method using the modulus quotient rule z2z1=∣z2∣∣z1∣:
∣1+i∣=12+12=2,∣1−i∣=12+(−1)2=2
1−i1+i=22=1
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- CBSE 2026Set ANNUAL1 markMCQQ.i9⋅i19 is equal to(a) −1(b) −i(c) 1(d) 0
›Reveal solutionSolution
Using i4=1, reduce the exponents mod 4: i9=i and i19=i3=−i, so the product is i⋅(−i)=1.
Recall i2=−1, and powers of i repeat with period 4: i1=i, i2=−1, i3=−i, i4=1.
i9=i4×2+1=(i4)2⋅i=1⋅i=i.
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- CBSE 2026Set ANNUAL1 markMCQQ.Conjugate of 3+5i is:(a) −3+5i(b) −3−5i(c) 3−5i(d) 3+i
›Reveal solutionSolution
The conjugate of z=a+ib is zˉ=a−ib; only the sign of the imaginary part changes.
Here z=3+5i, so a=3, b=5.
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- CBSE 2026Set ANNUAL1 markQ.Evaluate i9+i19.
›Reveal solutionSolution
Using i4=1, reduce each exponent mod 4: i9=i and i19=−i, which sum to 0.
Since i4=1, any power ik=i(kmod4).
i9=i(4×2+1)=(i4)2⋅i1=1⋅i=i.
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- CBSE 2026Set ANNUAL1 markQ.Write modulus of 5−3i.
›Reveal solutionSolution
Modulus of a+ib is a2+b2; here that gives 34.
For z=5−3i, a=5, b=−3.
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- CBSE 2026Set 1A1 markMCQQ.The value of −25×−9 is -(1) 15(2) −15(3) 15i(4) None of these
›Reveal solutionSolution
Write −25=5i and −9=3i; their product is 15i2=−15.
The rule ab=ab fails for negative numbers, so convert first:
−25=5i,−9=3i. …
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