Math Chapter Wise

EnglishPGT 2002

Q121.The locus represented by |z–1| = |z+i| is:

Aa circle of radius 1 Ban ellipse with foci at (1, 0) and (0,–1) CA straight line through the origine Da circle on the line joining (1,0), (0,1) as diameter

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EnglishPGT 2002

Q122.If z = x+iy and w= (1–iz)/(z–1), then |w|=1 implies that in the complex plane:

Az lies on the imaginary axis Bz lies on the real axis Cz lies on the unit circle DNone of these

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EnglishPGT 2002

Q123.The solution of the equation |z|–z=1 +2i is:

A3 2i 2 − B3 2i 2 + C−3i/2 DNone of these

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EnglishPGT 2002

Q124.The area of the triangle on the complex plane formed by the complex numbers z, iz and z+iz is:

A|z|2 B2/| z | C2 1 | z | 2 DNone of these

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EnglishPGT 2000

Q125.The loucs of point Z satisfying 1 Re K Z = is: (where k is non zero real number)

AA hyperbola BAn ellipse CA circle DA straight line

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EnglishPGT 2000

Q126.If 1+ω+ω2 = 0 and ω3 = 1 then the value of ω(1+ω–ω2) is:

A–2 B1 C4 D3

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EnglishPGT 2000

Q127.If 1 1 A iB i + = + + = + + = + + = + then the value B is:

A7 B3 C–1 D1

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EnglishRajasthan TGT 2016

Q128.If (((()))) 12 1 3i x iy + = + + = + + = + + = + then 1/12 y is equal to

A1 B2 C3 D0

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EnglishRajasthan TGT 2015

Q129.The amplitude of 3 i 3 i i 2 i 2 i − + + + − is equal to

A2/π B−2/π C−4/π D4/π

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EnglishRajasthan TGT 2015

Q130.If 4 z 2 z − = − = then the maximum value of z is

A5 1 B5 1/+ C5 DNone of these

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EnglishRajasthan TGT 2013

Q131.If z is complex number and + 1 z =a z, a > 2, then the greatest and least value of |z| are respectively-

A2 a a 4 2 + − and 2 a a 4 2 − + + B2 a a 4 2 + − and 2 a a 4 2 − − C2 a a 4 2 + + and 2 a a 4 2 − + + D2 a 4 + and 2 a 4 −

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EnglishRajasthan TGT 2013

Q132.The argument of x + iy, x - iy when y > 1, x is

A2 tan–1/y/x Bπ – 2 tan–1/y/x Ctan–1 2 2 x y − + D2 cot–1/y/x

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EnglishRajasthan TGT 2013

Q133.–7 – 24i =

A± (3 – 4i) B± (4 + 3i) C± (4 – 3i) D± (3 + 4i)

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EnglishRajasthan TGT 2011

Q134.The argument of complex number 1 i Z 1 i − = + is

A−2/π B2/π C3 4 π D5 2 π

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EnglishRajasthan TGT 2011

Q135.If complex numbers 1 2 Z,Z and 3 Z are vertices of an equilateral triangle, then 1 2 2 2 2 3 Z Z Z + + + + is equal to

A1 2 2 3 3 1 Z Z Z Z Z Z + + B1 2 2 3 3 1 2(Z Z Z Z Z Z) + + C1 2 2 3 3 1 1 (Z Z Z Z Z Z) 2 + + DNone of these

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EnglishJharkhand TGT 2017

Q136.The product of the value of 3 2 cos isin 6 6 π + is

Ai B–i C–1 D1

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EnglishJharkhand TGT 2017

Q137.For 'n' is an integer (cos θ–i sinθ)n =

A(cosθ+ i sinθ) Bn(cosθ– i sinθ) C(cos nθ–i sin nθ) Dcosθ–i sin θ

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EnglishJharkhand TGT 2017

Q138.(()) (()) 1 1 2 2 cos isin cos isin θ + θ + θ + θ + θ = θ + θ + θ + θ + θ

Acos (θ1+θ2)+i sin (θ1+θ2) B1 2 1 2/cos( ) isin( ) C1 D() 1 2 1 cos isin θ + θ

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EnglishJharkhand TGT 2017

Q139.What will be the argument of the complex number z=1-cosp+isinp?

Aπ-p Bp 4 2 π − Cp 2 2 π − Dp 2 π −

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EnglishJharkhand TGT 2017

Q140.If a complex number is given by z=(cosθ, sin θ), then (z-1/z)=

A(2cos θ,0) B(0, 0) C(0, 2i cos θ) D(0, 2i sin θ)

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