The number of values of z C, satisfying the equations |z-(4+8i)|= 10 and |z-(3+5i)|+|z-(5+11i)|=4 5, is :
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Previous year questions with solutions · 2024 – 2026
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 36 | 19 | 1.9 | 7.58% ▲ 2.9% vs last year |
| 2025 | 35 | 19 | 1.8 | 7.37% ▲ 0.5% vs last year |
| 2024 | 44 | 20 | 2.2 | 7.33% |
The number of values of z C, satisfying the equations |z-(4+8i)|= 10 and |z-(3+5i)|+|z-(5+11i)|=4 5, is :
Let z be a complex number such that |z+2|=|z-2| and ( z+3z-i )= 4. Then |z|2 is equal to:
Let the circles C1 : |z|=r and C2 : |z-3-4i|=5, z C, be such that C2 lies within C1. If z1 moves on C1, z2 moves on C2 and |z1-z2|=2, then |z1-z2| is equal to :
Let z be a complex number such that |z − 6| = 5 and |z + 2 − 6i| = 5. Then the value of z3 + 3z2 - 15z + 141 is equal to :
Let A = \z C : |z - 2| 4\ and B = \z C : |z - 2| + |z + 2| = 5\. Then the \|z1 - z2| : z1 A and z2 B\ is :
Let S = \z C : | z-6iz-2i | = 1 and | z-8+2iz+2i | = 35 \. Then z S |z|2 is equal to :
Let S = \z : 3 |2z - 3(1 + i)| 7 \ be a set of complex numbers. Then Minz S | (z + 12(5 + 3i) ) | is equal to :
Let z be the complex number satisfying |z - 5| 3 and having maximum positive principal argument. Then 34 | 5z - 125iz + 16 |2 is equal to:
Among the statements (S1) : The set \z C - \-i\ : |z| = 1 and z - iz + i is purely real \ contains exactly two elements, and (S2) : The set \z C - \-1\ : |z| = 1 and z - 1z + 1 is purely imaginary \ contains infinitely many elements.
If the locus of z C, such that Re ( z-12z+i ) + Re ( z-12 z-i ) = 2, is a circle of radius r and center (a, b), then 15abr2 is equal to :
If z1, z2, z3 C are the vertices of an equilateral triangle whose centroid is z0, then k=13 (zk - z0)2 is equal to
Let |z1 - 8 - 2i| 1 and |z2 - 2 + 6i| 2, z1, z2 C. Then the minimum value of |z1 - z2| is :
Let O be the origin, the point A be z1 = 3 + 2 2i, the point B(z2) be such that 3|z2| = |z1| and (z2) = (z1) + 6. Then
Let | z-i2 z+i |= 13, z C, be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points (0,0), C and ( ,0) is 11 square units, then 2 equals:
Let the curve z(1+i)+ z(1-i)=4, z , divide the region |z-3| 1 into two parts of areas and . Then | - | equals :
Let z be a complex number such that the real part of z - 2iz + 2i is zero. Then, the maximum value of |z - (6 + 8i)| is equal to
Let S1 = \z C : |z| 5\, S2 = \z C : Im ( z + 1 - 3 i1 - 3 i ) 0 \ and S3 = \z C : Re(z) 0\. Then the area of the region S1 S2 S3 is :
The area (in sq. units) of the region S = \z C : |z - 1| 2; (z + z) + i(z - z) 2, Im(z) 0\ is
Let S= \z :|z-1|=1 and ( 2-1)(z+ z)-i(z- z)=2 2 \. Let z1, z2 be such that |z1|= z S|z| and |z2|= z S|z|. Then | 2z1-z2 |2 equals :
If z is a complex number such that |z| 1, then the minimum value of |z+ 12(3+4 i) | is :