Text Solution
Verified by Experts
The correct Answer is:
Topper's Solved these Questions
OBJECTIVE SECTION AS PER NEW PAPER SCHEME
KUMAR PRAKASHAN|Exercise Linear in Equalities (Fill in the Blanks)|20 VideosOBJECTIVE SECTION AS PER NEW PAPER SCHEME
KUMAR PRAKASHAN|Exercise Linear in Equalities (True/False Statement)|14 VideosOBJECTIVE SECTION AS PER NEW PAPER SCHEME
KUMAR PRAKASHAN|Exercise Complex numbers and Quadratic Equations (Fill in the Blanks)|40 VideosMATHEMATICAL REASONING
KUMAR PRAKASHAN|Exercise QUESTION OF MODULE (Knowledge Test:)|2 VideosPERMUTATIONS AND COMBINATIONS
KUMAR PRAKASHAN|Exercise PRACTICE WORK |40 Videos
Similar Questions
Explore conceptually related problems
KUMAR PRAKASHAN-OBJECTIVE SECTION AS PER NEW PAPER SCHEME -Complex numbers and Quadratic Equations (True/False Statement)
- If (5,6) + z = (2,-1) then z = (3,-7)
Text Solution
|
- (0,2) . (0,2) = (0,4)
Text Solution
|
- i^(9)+i^(10)-3i^(12)=-4
Text Solution
|
- For complex number z =3 -2i,z + bar z = 2i Im(z) .
Text Solution
|
- For complex number z = 5 + 3i value z -barz=10.
Text Solution
|
- Inverse of (7,0) does not exists.
Text Solution
|
- The polar coordinate of complex number z=1+isqrt3 is (2,(pi)/3) .
Text Solution
|
- Equation having n degree has n solution.
Text Solution
|
- If z = x + iy and x +iy= (a+ib)/(a-ib) then x^2+y^2=1.
Text Solution
|
- If (5,6) + z = (2,-1) then z = (3,-7)
Text Solution
|
- (0,2) . (0,2) = (0,4)
Text Solution
|
- i^(9)+i^(10)-3i^(12)=-4
Text Solution
|
- For complex number z =3 -2i,z + bar z = 2i Im(z) .
Text Solution
|
- For complex number z = 5 + 3i value z -barz=10.
Text Solution
|
- Inverse of (7,0) does not exists.
Text Solution
|
- The polar coordinate of complex number z=1+isqrt3 is (2,(pi)/3) .
Text Solution
|
- Equation having n degree has n solution.
Text Solution
|
- If z = x + iy and x +iy= (a+ib)/(a-ib) then x^2+y^2=1.
Text Solution
|