Q.The value of −25×−9 is _____.
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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 …
Concept: Complex number arithmetic with square roots of negative numbers.
When dealing with square roots of negative numbers, we must work in the complex number system. The key principle is that −a=ia for any positive real number a, where i=−1.
Applying this rule to each factor:
−25=i25=5i
−9=i9=3i
Now multiply these results:
−25×−9=(5i)(3i)=15i2
Since i2=−1, we have:
15i2=15(−1)=−15 …
When multiplying square roots of negative numbers, convert each to imaginary form first: −25×−9=5i×3i=15i2=−15.
The trap here is tempting: you might want to write −25×−9=(−25)(−9)=225=15. That would be wrong. The rule a×b=ab holds only when at least one of a or b is non-negative. Once both are negative, we've left the real numbers and entered the complex plane, where the algebra of square roots changes.
The correct approach is to recognize that the square root of a negative number is an imaginary number. Recall that i=−1, so any −k for positive k can be written as k⋅i.
Step-by-step solution
- Convert each square root to imaginary form.
−25=25⋅(−1)=25⋅−1=5i
−9=9⋅(−1)=9⋅−1=3i …
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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