If $z\bar z = 0$, iff

${\mathop{\rm Re}\nolimits} (z) = 0$

${\mathop{\rm Im}\nolimits} (z) = 0$

Let ${z_1}$ and ${z_2}$ be two complex numbers with $\alpha $ and $\beta $ as their principle arguments such that ${\rm{\alpha }} + {\rm{\beta }} > {\rm{\pi }},$ then principal $\arg \left( {{z_1}{z_2}} \right)\;$ is given by

${\rm{\alpha }} + {\rm{\beta }} - {\rm{\pi }}$

${\rm{\alpha }} + {\rm{\beta }} + {\rm{\pi }}$

${\rm{\alpha }} + {\rm{\beta }} + 2{\rm{\pi }}$

${\rm{\alpha }} + {\rm{\beta }}$

Let a circle $C$ in complex plane pass through the points ${z_1} = 3 + 4i,\,\,{z_2} = 4 + 3i$ and $\,{z_3} = 5i.\,$ If $z( \ne {z_1})$ sis point on $C$ such that the line through $z$ and ${z_1}$ is perpendicular to the line through ${z_2}$ and ${z_3}$ then $\arg (z)$ is equal to:

${\tan ^{ - 1}}\left( {\frac{2}{{\sqrt 5 }}} \right) - \pi $

${\tan ^{ - 1}}\left( {\frac{3}{4}} \right) - \pi $

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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Easy

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121 Easy

121 of 612

Application +4 | -1

\sqrt {x + iy} = \pm \left( {a + ib} \right),

then

\sqrt { - x - iy}

is equal to

View answer and explanation

Correct option: C

\pm \left( {b - ia} \right)

Option C (

\pm \left( {b - ia} \right)

) is correct because it is derived by equating the real and imaginary parts of the given equation and then finding the square root of the negative conjugate.

122 Easy

122 of 612

Analyzing +4 | -1

Let

{z_1}\;\mathrm{and}\;{z_2}

be the non-real roots of the equation

3{z^2} + 3z + b = 0.

If the origin together with the points represented by

{z_1}

and

{z_2}

form an equilateral triangle, then the value of

b

View answer and explanation

Correct option: A

Option A (1) is correct because the condition for the origin and the two roots to form an equilateral triangle leads to a specific relationship between the roots and the coefficient b.

123 Easy

123 of 612

Understanding +4 | -1

z\bar z = 0

, iff

View answer and explanation

Correct option: C

(z) = 0

Option C (

(z) = 0

) is correct because

z\bar z = |z|^2 = 0

implies that the magnitude of z is 0, which means z itself is 0.

124 Easy

124 of 612

Application +4 | -1

x=\cos \alpha+i \sin \alpha

and

y=\cos \beta+\mathrm{i} \sin \beta

, then

(x-y) /

(x+y)

is equal to

View answer and explanation

Correct option: A

i \tan \left(\frac{\alpha-\beta}{2}\right)

Option A (

i \tan \left( {\frac{{\alpha - \beta}}{2}} \right)

) is correct because using trigonometric identities, the expression simplifies to

i \tan \left( {\frac{{\alpha - \beta}}{2}} \right)

125 Easy

125 of 612

Analyzing +4 | -1

Suppose

a \in \mathbf{R}

and the equation

z+a|z|+

2 i=0

has no solution in

\mathbf{C}

, then

a

satisfies the relation.

View answer and explanation

Correct option: B

|a| \geq 1

Option B (

|a| \geq 1

) is correct because the equation has no solution when the coefficient of |z|^2 is non-positive, which occurs when

|a| \geq 1

126 Easy

126 of 612

Application +4 | -1

\omega=e^{-2 \pi i / 3}, a, b, c, x, y, z

are non-zero complex numbers such that

a+b+c=x, a+b \omega+c \omega^2=y

and

a+

b \omega^2+c \omega=z

, then

\frac{|x|^2+|y|^2+|z|^2}{|a|^2+|b|^2+||^2}=

View answer and explanation

Correct option: D

Option D (9) is correct because, given the conditions and substituting a=b=c, x=3a, y=0, z=0, the expression simplifies to 9.

127 Easy

127 of 612

Application +4 | -1

{7^{{{\log }_7}\left( {{x^2} - 4\;x + 5} \right)}} = x - 1,x

may have values

View answer and explanation

Correct option: A

2,\;3

Option A (2, 3) is correct because these values satisfy the given equation when substituted and checked for validity within the logarithmic function.

128 Easy

128 of 612

Application +4 | -1

The value of

(1+i)^5(1-i)^5

2^n,

n

' is equal to

View answer and explanation

Correct option: A

Option A (5) is correct because

(1+i)^5(1-i)^5

simplifies to

2^5

, thus n=5.

129 Easy

129 of 612

Application +4 | -1

The value of

\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1

View answer and explanation

Correct option: B

-2

Option C (-2) is correct because after simplification, the expression equals -2.

130 Easy

130 of 612

Application +4 | -1

z_1=2+3 i

and

z_2=3-2 i

, then

z_1-z_2

equals to

-1+ bi.

The value of '

b

' is

View answer and explanation

Correct option: A

Option A (5) is correct because subtracting z2 from z1 results in -1 + 5i, so the value of 'b' is 5.

131 Easy

131 of 612

Understanding +4 | -1

z \neq 0

, then

\int\limits_{{x=0}}^{100}[\arg |z|] d x

is (where

[*]

denotes the greatest integer function)

View answer and explanation

Correct option: A

Option A is correct because the argument of the absolute value of any non-zero complex number is always 0, and the greatest integer function of 0 is 0.

132 Easy

132 of 612

Application +4 | -1

The mirror image of the curve

\arg \left(\frac{z+i}{z-1}\right)=\frac{\pi}{4}

i=\sqrt{-1}

in the line

x-y=0

, is

View answer and explanation

Correct option: C

\arg \left(\frac{z-i}{z+1}\right)=\frac{\pi}{4}

Option C (

\arg \left(\frac{{z - i}}{{z + 1}}\right) = \frac{\pi }{4}

) is correct because the mirror image of a complex number z in the line x-y=0 is obtained by swapping the real and imaginary parts, which is equivalent to taking

i\bar{z}

133 Easy

133 of 612

Application +4 | -1

Let

z

be a complex number such that

\left| {\frac{{z - i}}{{z + 2i}}} \right| = 1

and

\left| z \right| = \frac{5}{2}.

Then the value of

\left| {z + 3i} \right|

View answer and explanation

Correct option: B

\frac{7}{2}

Option B (7/2) is correct because, after finding the locus of z and using the given modulus condition, we can determine |z + 3i| = 7/2.

134 Easy

134 of 612

Application +4 | -1

|z|=z+3-2 i

, then

z

equals

View answer and explanation

Correct option: C

-5 / 6+2 i

Option C (-5/6 + 2i) is correct because solving the equation |z| = z + 3 - 2i for z yields z = -5/6 + 2i.

135 Easy

135 of 612

Application +4 | -1

{z_1} = \sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)

And

{z_2} = \sqrt 3 \left( {\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}} \right)

, then

\left| {{z_1}{z_2}} \right|

View answer and explanation

Correct option: C

\sqrt 6

The correct answer is C (

\sqrt 6

) because the modulus of the product of two complex numbers is the product of their moduli.

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136 Easy

136 of 612

Analyzing +4 | -1

Let

z\left( { \ne 2} \right)

be a complex number such that

{\log _{1/2}}\left| {z - 2} \right| > {\log _{1/2}}\left| z \right|,

then

View answer and explanation

Correct option: A

\mathrm{Re\;}\left( z \right) > 1

Option A (

\mathrm{Re\;}\left( z \right) > 1

) is correct because the inequality implies that the distance from z to 2 is less than the distance from z to 0, meaning z lies to the right of the perpendicular bisector of the segment joining 0 and 2.

137 Easy

137 of 612

Application +4 | -1

The conjugate of complex number

\frac{{2 - 3i}}{{4 - i}}

View answer and explanation

Correct option: B

\frac{{11 + 10i}}{{17}}

Option B (

\frac{{11 + 10i}}{{17}}

) is correct because it is the result of finding the conjugate of the given complex number and simplifying the expression.

138 Easy

138 of 612

Application +4 | -1

\omega

is a complex cube root of unity, then

225 + {\left( {3\omega + 8{\omega ^2}} \right)^2} + {\left( {3{\omega ^2} + 8\omega } \right)^2}

is equal to

View answer and explanation

Correct option: D

248

Option D (248) is correct because it is the result of simplifying the expression using the properties of complex cube roots of unity.

139 Easy

139 of 612

Application +4 | -1

Let

z

be a complex number such that

\left| z \right| + z = 3 + i

(where

i = \sqrt { - 1} )

. Then

\left| z \right|

is equal to

View answer and explanation

Correct option: B

\frac{5}{3}

Option B (5/3) is correct because by substituting

z = x + i

into the given equation and solving for x, we can find the magnitude of z.

140 Easy

140 of 612

Understanding +4 | -1

Let

{z_1}

and

{z_2}

be two complex numbers with

\alpha

and

\beta

as their principle arguments such that

\mathrm{\alpha } + \mathrm{\beta } > \mathrm{\pi },

then principal

\arg \left( {{z_1}{z_2}} \right)\;

is given by

View answer and explanation

Correct option: B

\mathrm{\alpha } + \mathrm{\beta } - \mathrm{\pi }

Option B (

\alpha + \beta - \pi

) is correct because the principal argument must lie between

-\pi

and

\pi

, and since

\alpha + \beta > \pi

, we subtract

2\pi

to bring it into the range.

141 Easy

141 of 612

Understanding +4 | -1

The complex numbers

{z_1},{z_2},{z_3},{z_4}

taken in that order in the Argand plane represent the vertices of a parallelogram iff

View answer and explanation

Correct option: B

{z_1} + {z_3} = {z_2} + {z_4}

Option B (

{z_1} + {z_3} = {z_2} + {z_4}

) is correct because in a parallelogram, the sum of opposite vertices are equal.

142 Easy

142 of 612

Application +4 | -1

1,\;\omega ,\;{\omega ^2}

are the cube roots of unity, then

{\left( {3 + {\omega ^2} + {\omega ^4}} \right)^6}

is equal to

View answer and explanation

Correct option: A

Option A (64) is correct because simplifying the expression using the properties of cube roots of unity results in 2⁶ = 64.

143 Easy

143 of 612

Analyzing +4 | -1

Let a circle

C

in complex plane pass through the points

{z_1} = 3 + 4i,\,\,{z_2} = 4 + 3i

and

\,{z_3} = 5i.\,

z( \ne {z_1})

sis point on

C

such that the line through

z

and

{z_1}

is perpendicular to the line through

{z_2}

and

{z_3}

then

\arg (z)

is equal to:

View answer and explanation

Correct option: B

{\tan ^{ - 1}}\left( {\frac{{24}}{7}} \right) - \pi

The correct answer is B (

{\tan ^{ - 1}}\left( {\frac{{24}}{7}} \right) - \pi

) because it is the argument of the point z on the circle, calculated using the given conditions and complex number properties.

144 Easy

144 of 612

Application +4 | -1

\omega

and

{\omega ^2}

are the two imaginary cube roots of unity, then the equation whose roots are

a{\omega ^{317}}

and

a{\omega ^{382}}

, is

View answer and explanation

Correct option: A

{x^2} + ax + {a^2} = 0

Option A (

{x^2} + ax + {a^2} = 0

) is correct because it is the quadratic equation formed using the given roots and properties of cube roots of unity.

145 Easy

145 of 612

Understanding +4 | -1

arg

\left( {\bar z} \right)

is equal to

View answer and explanation

Correct option: B

2\mathrm{\pi } - \arg \left( z \right)

Option B (2π - arg(z)) is correct because the argument of the conjugate of a complex number is the negative of the argument of the original number, or 2π minus the argument.

146 Easy

146 of 612

Application +4 | -1

{\left( {\frac{{ - 1 + \sqrt { - 3} }}{2}} \right)^{100}} + {\left( {\frac{{ - 1 - \sqrt { - 3} }}{2}} \right)^{100}}

is equal to

View answer and explanation

Correct option: C

- 1

Option C (-1) is correct because the expression simplifies to

\omega + {\omega ^2}

which equals -1, where

\omega

is a complex cube root of unity.

147 Easy

147 of 612

Application +4 | -1

Let

{z_1}

be a complex number with

\left| {{z_1}} \right| = 1

and

{z_2}

be any complex number, then

\left| {\frac{{{z_1} - {z_2}}}{{1 - {z_1}{z_2}}}} \right|

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because the modulus of the given expression simplifies to 1 using the property that |z1| = 1.

148 Easy

148 of 612

Analyzing +4 | -1

\cos A + \cos B + \cos C = 0,

\sin A + \sin B + \sin C = 0

and

A + B + C = 180^\circ ,

then the value of

\cos 3\;A + \cos 3\;B + \cos 3\;C

View answer and explanation

Correct option: B

- 3

Option C (1/3) is correct because by substituting a = 0 and b = 1, we can solve for x and y and then find the value of 1/x - 1/y.

149 Easy

149 of 612

Application +4 | -1

\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{48}=3^{24}(x+i y)

, then the ordered pair

(x, y)

View answer and explanation

Correct option: C

(1,0)

Option C (

(1,0)

) is correct because it is the simplified form of the given complex number expression using De Moivre's theorem.

150 Easy

150 of 612

Application +4 | -1

z

is a complex number such that

-\pi / 2 \leq

\arg z \leq \pi / 2

, then which of the following inequality is true.

View answer and explanation

Correct option: A

|z-\bar{z}| \leq|z||\arg (z)-\arg (\bar{z})|

Option A (

|z-\bar{z}| \leq |z||\arg (z)-\arg (\bar{z})|

) is correct because it follows from the geometric interpretation of complex numbers and their arguments.

Revision notes and key takeaways

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