Q.Express the following in the form a+ib: (−2−31i)3
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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 …
Concept: Complex Number Arithmetic — expanding a binomial cube with a complex number.
First, write the expression as −(2+31i) and cube it:
(−2−31i)3=(−1)3(2+31i)3=−(2+31i)3.
Now expand (2+31i)3 using (p+q)3=p3+3p2q+3pq2+q3:
23+3(2)2(31i)+3(2)(31i)2+(31i)3.
Compute term by term:
- 23=8
- 3⋅4⋅31i=4i
- 3⋅2⋅91i2=6⋅91(−1)=−32
- 271i3=271(−i)=−271i
So the sum is:
This problem uses the binomial expansion of a complex cube. The key is to treat (−2−31i) as a binomial, expand carefully with powers of i, and combine real and imaginary parts. The final result is −322−27107i.
We start with a complex number in the form x+iy, but here x=−2 and y=−31. Cubing it directly means multiplying it by itself three times. That’s doable but messy. A cleaner way is to use the binomial theorem: (a+b)3=a3+3a2b+3ab2+b3. Here, a=−2 and b=−31i. The trick is to keep track of the powers of i — they cycle, and that’s where students often slip.
Let’s work through it step by step.
- Write the expression and set up the binomial expansion. We have (−2−31i)3. Let a=−2 and b=−31i. Then
(a+b)3=a3+3a2b+3ab2+b3.
-
Compute each term separately.
- a3=(−2)3=−8.
- 3a2b=3⋅(−2)2⋅(−31i)=3⋅4⋅(−31i)=12⋅(−31i)=−4i.
- 3ab2=3⋅(−2)⋅(−31i)2. First square b: (−31i)2=91i2=91(−1)=−91. Then 3ab2=3⋅(−2)⋅(−91)=−6⋅(−91)=96=32.
- b3=(−31i)3=(−31)3⋅i3=−271⋅i3. Now i3=i2⋅i=(−1)⋅i=−i, so b3=−271⋅(−i)=271i.
-
Combine all terms.
Sum them: (−8)+(−4i)+(32)+(271i).
Group the real parts: −8+32=−324+32=−322. …
Showing the 12 most recent of 59 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.If the complex number z=x+iy satisfies the equation 5z−2zˉ=1+i7−7i, then the value of x+y is equal to (A) 7 (B) -3 (C) 3 (D) -1 (E) -7
›Reveal solutionSolution
Simplify the right side to −7i, equate real and imaginary parts, get x=0,y=−1, so x+y=−1.
Simplify the right-hand side:
1+i7−7i=(1+i)(1−i)(7−7i)(1−i)=27−7i−7i+7i2=2−14i=−7i.
With z=x+iy, zˉ=x−iy:
5z−2zˉ=5(x+iy)−2(x−iy)=3x+7iy. …
- KEAM 2026Set eng-2026-04174 marksMCQQ.Let z=1+i, where i=−1. If z−z224zˉ=λz, then the value of λ is equal to (A) 12 (B) 13 (C) 18 (D) 23 (E) 24
›Reveal solutionSolution
Evaluate z224zˉ=−12−12i; subtracting from z gives 13(1+i)=13z, so λ=13.
For z=1+i: z2=(1+i)2=2i and zˉ=1−i. Then
z224zˉ=2i24(1−i)=i12(1−i)=12(1−i)(−i)=12(−i+i2)=−12−12i.
Therefore …
- KEAM 2026Set eng-2026-04174 marksMCQQ.The complex number z satisfying the equation 2+iRe(z)+1+2iIm(z)=1−2i3, is (A) 6−5i (B) −4+5i (C) 5−4i (D) 3−5i (E) 4−5i
›Reveal solutionSolution
Rationalize each term and match real/imaginary parts to solve 2a+b=3, −a−2b=6, giving z=4−5i.
Let Re(z)=a, Im(z)=b (both real). Rationalizing:
2+ia=5a(2−i),1+2ib=5b(1−2i),1−2i3=53(1+2i).
Multiplying through by 5:
a(2−i)+b(1−2i)=3(1+2i). …
- KEAM 2026Set eng-2026-04184 marksMCQQ.Let z=i−2a−2i, where a is a real number and i=−1. If Im(z)=0, then the value of a is equal to (A) 1 (B) 2 (C) 3 (D) 4 (E) 0
›Reveal solutionSolution
Multiply by the conjugate; setting the imaginary part to zero gives a=1.
z=i−2a−2i. Multiply numerator and denominator by (i−2)=(−2−i); denominator =(−2)2+12=5.
Numerator:
(a−2i)(−2−i)=(−2a−21)+i(1−a). …
- KEAM 2026Set eng-2026-04184 marksMCQQ.The value of [(3+i)(3−i)5i]2026 is equal to (A) 220261 (B) 210131 (C) 21013−1 (D) 22026i (E) 22026−1
›Reveal solutionSolution
Base simplifies to i/2; i2026=−1, giving 22026−1.
(3+i)(3−i)=9−i2=10, so
(3+i)(3−i)5i=105i=2i.
Then
(2i)2026=22026i2026. …
- KEAM 2026Set eng-2026-04184 marksMCQQ.If z∣z∣=24+7i, where z is a complex number, then the value of ∣z∣ is equal to (A) 5 (B) 7 (C) 12 (D) 15 (E) 25
›Reveal solutionSolution
∣z∣⋅∣z∣=∣24+7i∣=25⇒∣z∣=5.
Take the modulus of both sides of z∣z∣=24+7i. Since ∣z∣ is real and positive, …
- KEAM 2026Set eng-2026-04194 marksMCQQ.Let z1=7−5i5+7i,z2=3−2i3+2i and z3=11−i1+11i. Then z1z1+z2z2+z3z3 is equal to (A) 2 (B) 1+2i (C) 1 (D) 3 (E) 1−2i
›Reveal solutionSolution
zz=∣z∣2, and each fraction has numerator and denominator of equal modulus.
For a complex number, zz=∣z∣2, and qp2=∣q∣2∣p∣2.
- z1=7−5i5+7i: 49+2525+49=7474=1. …
- KEAM 2026Set eng-2026-04204 marksMCQQ.Given that i2=−1. If z1=(7+i5)2+(7−i5)2 and z2=(3+2i)3−(3−2i)3, then (A) z1 is a purely imaginary number and z2 is a purely real number (B) z1 is a purely real number and z2 is a purely imaginary number (C) both z1 and z2 are purely imaginary numbers (D) both z1 and z2 are purely real numbers (E) z1+z2 is a purely real number
›Reveal solutionSolution
A number plus its conjugate is real; a number minus its conjugate is imaginary. z1 is a sum of conjugates so it is purely real, while z2 is a difference of conjugates so it is purely imaginary.
Let w=(7+i5)2. Then (7−i5)2=w, so
z1=w+w=2Re(w),
which is a purely real number.
(Explicitly w=49−5+14i5=44+14i5, so z1=88.)
Let u=(3+2i)3. Then (3−2i)3=u, so …
- KEAM 2026Set eng-2026-04204 marksMCQQ.If z1=1+3i, z2=−3i+5, then (z1z2+z2z1)+(z1z2+z2z1) is equal to (A) −16 (B) 1+i (C) 1 (D) 1−i (E) −16i
›Reveal solutionSolution
The expression z1z2+z2z1 equals 2Re(z1z2)=−8, which is real. Adding its own conjugate (also −8) gives −16.
With z1=1+3i and z2=5−3i, so z2=5+3i.
Compute z1z2=(1+3i)(5+3i)=5+3i+15i+9i2=5+18i−9=−4+18i.
Note z2z1=z1z2=−4−18i, so
z1z2+z2z1=(−4+18i)+(−4−18i)=−8, …
- KEAM 2026Set eng-2026-04214 marksMCQQ.If ∣z+4∣=2∣z+1∣, where z is a complex number, then ∣z∣ is equal to (A) 0 (B) 2 (C) 4 (D) 8 (E) 16
›Reveal solutionSolution
Squaring the modulus condition gives the circle x2+y2=4, so ∣z∣=2.
Let z=x+iy. Then ∣z+4∣=2∣z+1∣ squared:
(x+4)2+y2=4[(x+1)2+y2]. …
- KEAM 2026Set eng-2026-04214 marksMCQQ.If z(3−i)=2+i, then z2= (A) 2i (B) 2−i (C) 21 (D) 2−1 (E) 21+i
›Reveal solutionSolution
z=21+i and squaring gives 2i.
From z(3−i)=2+i:
z=3−i2+i=(3−i)(3+i)(2+i)(3+i)=106+2i+3i−1=105+5i=21+i.
Then …
- KEAM 2026Set eng-2026-04214 marksMCQQ.The imaginary part of 1+i31−i3 is (A) 2−1 (B) 21 (C) 23 (D) 2−3 (E) 43
›Reveal solutionSolution
Rationalizing gives 2−1−i3, whose imaginary part is −23.
Multiply numerator and denominator by the conjugate 1−i3: …
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