datavorous/entrance-exam-dataset huggingface dataset:
If $\mathrm{z}_1, \mathrm{z}_2$ and $\mathrm{z}_3, \mathrm{z}_4$ are 2 pairs of complex conjugate numbers, then $\arg \left(\frac{z_1}{z_4}\right)+\arg \left(\frac{z_2}{z_3}\right)$ equals:

Options:
A
0

B
$\frac{\pi}{2}$

C
$\frac{3 \pi}{2}$

D
$\pi$

The correct answer is:
0

Solution:
Consider $\arg \left(\frac{z_1}{z_4}\right)+\arg \left(\frac{z_2}{z_3}\right)$
$$
=\arg \left(z_1\right)-\arg \left(z_4\right)+\arg \left(z_2\right)-\arg \left(z_3\right)
$$
$$
\begin{aligned}
&=\left(\arg \left(z_1\right)+\arg \left(z_2\right)\right)-\left(\arg \left(z_3\right)+\arg \left(z_4\right)\right) \\
&\quad \operatorname{given}\left(\begin{array}{l}
z_2=\bar{z}_1 \ \\
z_4=\bar{z}_3
\end{array}\right) \\
&=\left(\arg \left(z_1\right)+\arg \left(\bar{z}_1\right)\right)-\left(\arg \left(z_3\right)+\arg \left(\bar{z}_3\right)\right) \\
&\qquad\left\{\begin{array}{l}
\operatorname{also}\left(\arg \left(\bar{z}_1\right)=-\arg \left(z_1\right)\right. \\
\arg \left(\bar{z}_3\right)=-\arg \left(z_3\right)
\end{array}\right\} \\
&=\left(\arg \left(z_1\right)-\arg \left(z_1\right)\right)-\left(\arg \left(z_3\right)-\arg \left(z_3\right)\right) \\
&=0-0=0
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2014 (11 Apr Online)
If z1, z2 are complex numbers such that Rez1=z1-1 and Rez2=z2-1 and argz1-z2=π6, then Imz1+z2 is equal to :

Options:
A 23
B 32
C 13
D 23

The correct answer is:
23

Solution:
Let z1 =x1+iy1 and z2=x2+iy2

Given z1-1=Rez1

⇒x1-12+y12=x12

⇒y12-2x1+1=0 ..............1

z2-1=Rez2

⇒x2-12+y22=x22

⇒y22 -2x2+1=0 ...........2

Subtracting equation 2 from equation 1, we get

y12-y22-2x1-x2=0

⇒y1-y2 y1+y2=2x1-x2

⇒y1+y2=2x1-x2y1-y2 ............3

Also given, arg z1-z2=π6

⇒tan-1y1-y2x1-x2=π6

⇒y1-y2x1-x2=13 .........4

Putting equation 4 in equation 3, we get

y1+y2=23⇒Imz1+z2=23

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If $z_1, z_2$ are two complex numbers satisfying $\left|\frac{z_1-3 z_2}{3-z_1 \bar{z}_2}\right|=1,\left|z_1\right| \neq 3$, then $\left|z_2\right|$ is equal to

Options:
A $1$
B $2$
C $3$
D $4$

The correct answer is:
$1$

Solution:
$\begin{aligned} & \text { Given that }\left|\frac{z_1-3 z_2}{3-z_1 \bar{z}_2}\right|=1,\left|z_1\right| \neq 3 \\ & \Rightarrow\left|z_1-3 z_2\right|=\left|3-z_1 \bar{z}_2\right| \quad\left[\because\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\right] \\ & \Rightarrow \quad\left|z_1-3 z_2\right|^2=\left|3-z_1 \bar{z}_2\right|^2 \\ & \Rightarrow \quad\left(z_1-3 z_2\right)\left(\bar{z}_1-3 \bar{z}_2\right)=\left(3-z_1 \bar{z}_2\right)\left(3-\bar{z}_1 z_2\right) \\ & {\left[\because \bar{z}=z_2\right]} \\ & \Rightarrow \quad\left|z_1\right|^2-3 z_1 \bar{z}_2-3 z_2 \bar{z}_1+9\left|z_2\right|^2 \\ & =9-3 \bar{z}_1 z_2-3 z_1 \bar{z}_2+\left|z_1\right|^2\left|z_2\right|^2 \\ & \Rightarrow \quad\left|z_1\right|^2+9\left|z_2\right|^2-9-\left|z_1\right|^2\left|z_2\right|^2=0 \\ & \Rightarrow \quad\left(9-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right)=0 \\ & \Rightarrow \quad\left|z_1\right|^2=9 \text { or }\left|z_2\right|^2=1 \\ & \Rightarrow \quad\left|z_1\right|=3 \mid \text { or }\left|z_2\right|=1 \\ & \end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If z1,z2,z3 are 3 distinct complex such that 3z1-z2=5z2-z3=7z3-z1, then the value of 9z¯3z1-z2+25z¯1z2-z3+49z¯2z3-z1 is equal to

Options:
A 0
B 1
C -1
D 15

The correct answer is:
0

Solution:
Let, 3z1-z2=5z2-z3=7z3-z1=k
9z1-z22=25z2-z32=49z3-z12=k2
⇒9z1-z2z-1-z-2=k2⇒9z1-z2=k2z-1-z-2
Similarly, 25z2-z3=k2z¯3-z¯1
And, 49z3-z1=k2z¯3-z¯1
⇒9z¯3z1-z2+25z¯1z2-z3+49z¯2z3-z1=k2z¯1z¯3-z¯2z¯3+z¯2z¯1-z¯3z¯1+z¯3z¯2-z¯1z¯2=0

Topic: Mathematics, Complex Number, JEE Main
If $z_1, z_2 z_3, z_4$ are the roots of the equation $z^4=1$, then the value of $\sum_{i=1}^4 z_i^3$ is

Options:
A $0$
B $1$
C i
D $1+i$

The correct answer is:
$0$

Solution:
$\begin{aligned} & z_1=1, z_2=i, z_3=-1, z_4=-i \\ & \therefore 1^3+(i)^3+(-1)^3+(-i)^3=0\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\left|z_{1}\right|=\left|z_{2}\right|=\ldots \ldots \ldots . .\left|z_{n}\right|=1,$ then the

value of $\left|z_{1}+z_{2}+\ldots \ldots . z_{n}\right|-$

$\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots \ldots .+\frac{1}{z_{n}}\right|$ is,

Options:
A 0
B 1
C -1
D None

The correct answer is:
0

Solution:
$z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2}=\ldots .=z_{n} \bar{z}_{n}=1$

$\Rightarrow \overline{\mathrm{z}}_{1}=\frac{1}{\mathrm{z}_{1}}, \overline{\mathrm{z}}_{2}=\frac{1}{\mathrm{z}_{2}}, \overline{\mathrm{z}}_{3}=\frac{1}{\mathrm{z}_{3}}, \ldots \ldots, \overline{\mathrm{z}}_{\mathrm{n}}=\frac{1}{\mathrm{z}_{\mathrm{n}}}$

$\therefore\left|z_{1}+z_{2}+\ldots .+z_{n}\right|-\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots .+\frac{1}{z_{n}}\right|$

$=\left|z_{1}+z_{2}+\ldots+z_{n}\right|-\left|\bar{z}_{1}+\bar{z}_{2}+\ldots .+\bar{z}_{n}\right|=0$

Topic: Mathematics, Complex Number, JEE Main
If $\left|z^2-1\right|=|z|^2+1$, then $z$ lies on

Options:
A
the real axis

B
an ellipse

C
a circle

D
the imaginary axis.

The correct answer is:
the imaginary axis.

Solution:
$\left|z^2-1\right|^2=\left(|z|^2+1\right) \Rightarrow\left(z^2-1\right)\left(\bar{z}^2-1\right)=|z|^4+2|z|^2+1$
$\Rightarrow z^2+\bar{z}^2+2 z \bar{z}=0 \Rightarrow z+\bar{z}=0$
$\Rightarrow R(z)=0 \Rightarrow z$ lies on the imaginary axis.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2004
If Z-2=2Z-1, then the value of ReZZ2 is (where Z is a complex number and Re(Z) represents the real part of Z). If your answer is K, mark 4K as answer after rounding off to nearest integer.

Options:

The correct answer is:
3

Solution:
Let, Z=x+iy

⇒x-2+iy=2x-1+iy
⇒x-22+y2=4x-12+y2
⇒3x2+y2-4x=0
⇒xx2+y2=34
⇒ReZZ2=0.75

Topic: Mathematics, Complex Number, JEE Main
If Z-2=2Z-1, then the value of ReZZ2 is (where Z is a complex number and Re(Z) represents the real part of Z)

Options:

The correct answer is:
0.75

Solution:
Let, Z=x+iy

⇒x-2+iy=2x-1+iy
⇒x-22+y2=4x-12+y2
⇒3x2+y2-4x=0
⇒xx2+y2=34
⇒ReZZ2=0.75

Topic: Mathematics, Complex Number, JEE Main
If $z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5$ then

Options:
A $\operatorname{Re}(z)=0$
B $\operatorname{Im}(z)=0$
C $\operatorname{Re}(z)\gt0, \operatorname{Im}(z)\gt0$
D $\operatorname{Re}(z)\gt0, \operatorname{Im}(z) \lt 0$

The correct answer is:
$\operatorname{Im}(z)=0$

Solution:
Given that $z=\left(\frac{\sqrt{3}}{2}+i \frac{1}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-i \frac{1}{2}\right)^5$
$\begin{aligned} & =\left[\cos \left(\frac{\pi}{6}\right)+i \sin \left(\frac{\pi}{6}\right)\right]^5+\left[\cos \left(\frac{\pi}{6}\right)-i \sin \left(\frac{\pi}{6}\right)\right]^5 \\ & =\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}+\cos \frac{5 \pi}{6}-i \sin \frac{5 \pi}{6}\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If z=2+3i, then z5+z¯5 is equal to:

Options:
A 244
B 224
C 245
D 265

The correct answer is:
244

Solution:
Given z=2+3i,

So, z5+z¯5=2+3i5+2-3i5

Applying binomial expansion we get,

=2C0525+C25233i2+C45213i4

=232+10×8-9+5×2×81=244

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (29 Jul Shift 1)
If $z^2+z+1=0$, where $z$ is a complex number, then the value of
$$
\left(z+\frac{1}{z}\right)^2+\left(z^2+\frac{1}{z^2}\right)^2+\left(z^3+\frac{1}{z^3}\right)^2+\cdots+\left(z^6+\frac{1}{z^6}\right)^2
$$

Options:
A
18

B
54

C
6

D
12

The correct answer is:
12

Solution:
$z^2+z+1=0 \quad \Rightarrow z=\omega$ or $\omega^2$ so, $z+\frac{1}{z}=\omega+\omega^2=-1, z^2+\frac{1}{z^2}=\omega^2+\omega=-1, z^3+\frac{1}{z^3}=\omega^3+\omega^3=2$ $z^4+\frac{1}{z^4}=-1, z^5+\frac{1}{z^5}=-1$ and $z^6+\frac{1}{z^6}=2$ $\therefore$ The given sum $=1+1+4+1+1+4=12$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2006
If z2+z+1=0,z∈C, then ∑n=115zn+(-1)a1zn2 is equal to _____.

Options:

The correct answer is:
2

Solution:
Given, z2+z+1=0⇒z=w,w2

Now, ∑n=115zn+-11zn2=∑n=115z2n+1z2n+2-1n

=∑n=115w2n+1w2n+2-1n

=w21-w301-w2+1w21-1w301-1w2+2-1

=w21-11-w2+1w21-11-1w2-2

=0+0-2=2

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (26 Jun Shift 2)
If $\omega=\frac{z}{z-\frac{1}{3} i}$ and $|\omega|=1$, then $z$ lies on

Options:
A
an ellipse

B
a circle

C
a straight line

D
a parabola.

The correct answer is:
a straight line

Solution:
As given $w=\frac{z}{z-\frac{1}{3} i} \Rightarrow|w|=\frac{|z|}{\left|z-\frac{1}{3} i\right|}=1 \Rightarrow$ distance of $z$ from origin and point
$\left(0, \frac{1}{3}\right)$ is same hence $z$ lies on bisector of the line joining points $(0,0)$ and $(0,1 / 3)$.
Hence $z$ lies on a straight line.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2005
If z=32+i2 i=-1, then 1+iz+z5+iz89 is equal to:

Options:
A -1
B 1
C 0
D -1+2i9

The correct answer is:
-1

Solution:
z=32+i2 ∴iz=-12+32i=ω
∴1+iz+z5+iz89=1+ω+ω5i5+i×ω8i89
=1+ω-iω2+iω29
=-ω29=-ω18
= -1

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (08 Apr Shift 2)
If $|z-3+2 i| \leq 4$ then the difference between the greatest value and the least value of $|z|$ is

Options:
A
$\sqrt{13}$

B
$2 \sqrt{13}$

C
8

D
$4+\sqrt{13}$

The correct answer is:
$2 \sqrt{13}$

Solution:
$|z-(3-2 i)| \leq 4$ represents a circle whose centre is $(3,-2)$ and radius $=4$.
$|z|=|z-0|$ represents the distance of point ' $z$ ' from origin $(0,0)$
Suppose $R S$ is the normal of the circle passing through origin ' $O$ ' and $G$ is its center $(3,-2)$
Here, $O R$ is the least distance and $O S$ is the greatest distance $O R=R G-O G$ and $O S=O G+G S \ldots(1)$ As, $R G=G S=4$ $O G=\sqrt{3^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}$ From (1), $O R=4-\sqrt{13}$ and $O S=4+\sqrt{13}$ So, required difference
$$
\begin{aligned}
&=(4+\sqrt{13})-(4-\sqrt{13}) \\
&=\sqrt{13}+\sqrt{13}=2 \sqrt{13}
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
If $z=\frac{7-i}{3-4 i}$ then $z^{14}=$

Options:
A $2^7$
B $2^7 i$
C $2^{14} i$
D $-2^7 i$

The correct answer is:
$-2^7 i$

Solution:
$\begin{aligned} & z=\frac{7-i}{3-4 i} \times \frac{3+4 i}{3+4 i}=\frac{21+25 i+4}{16+9}=\frac{25(1+i)}{25}=(1+i) \\ & z^{14}=(1+i)^{14}=\left[(1+i)^2\right]^7=(2 i)^7=2^7 i^7=-2^7 i .\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If z=3â4i , then the value of expression z 4 â3 z 3 +3 z 2 +99zâ95 is equal to

Options:
A 5
B 6
C −5
D −4

The correct answer is:
5

Solution:
z=3â4i â (zâ3) 2 =â16
â z 2 â6z+25=0

âz4â3z3+3z2+99zâ95 =( z 2 +3zâ4)( z 2 â6z+25)+5 =( z 2 +3zâ4)(0)+5=5

Topic: Mathematics, Complex Number, JEE Main
If $|z+4| \leq 3$, then the maximum value of $|z+1|$ is

Options:
A
$4$

B
$10$

C
$6$

D
$0$

The correct answer is:
$6$

Solution:
From the Argand diagram maximum value of $|z+1|$ is $6$.
Alternative:
$\begin{aligned}
& |z+1|=|z+4-3| \\
& \leq|z+4|+|-3|=6.
\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2007
If $|z-4| < |z-2|$, its solution is given by

Options:
A
Re(z) > 0

B
Re(z) < 0
C
Re (z) > 3

D
Re(z) > 2

The correct answer is:
Re (z) > 3

Solution:
Given $|z-4| < |z-2|$ Let $z=x+$ iy
$$
\begin{aligned}
& \Rightarrow \mid(x-4)+i y)| < |(x-2)+i y \mid \Rightarrow(x-4)^2+y^2 < (x-2)^2+y^2 \\
& \Rightarrow x^2-8 x+16 < x^2-4 x+4 \Rightarrow 12 < 4 x \Rightarrow x>3 \Rightarrow \operatorname{Re}(z)>3
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2002
If z=a+ib satisfies arg(z-1)=arg(z+3i), then (a-1):b=

Options:
A 2:1
B 1:3
C -1:3
D None of these

The correct answer is:
1:3

Solution:
We have,

arg(z-1)=arg(z+3i)

⇒arg((a-1)+ib)=arg(a+(b+3)i)

⇒tan-1ba-1=tan-1b+3a

⇒ ba-1=b+3a

⇒ ab=(a-1)(b+3)

⇒ 3(a-1)=b

⇒ (a-1):b=1:3

Topic: Mathematics, Complex Number, JEE Main
If Z=Asin⁡θ+Bcos⁡θA+B; then

Options:
A the dimensions of Z and A are the same
B the dimensions of Z and B are the same
C Z is dimensionless quantity
D none of these

The correct answer is:
Z is dimensionless quantity

Solution:
Z=Asin⁡θ+Bcos⁡θA+B; Dimensions of A and B would be same as two quantities can be added only when they have same dimensions. So from given equation, we can say that Z is dimensionless.

Topic: Physics, Units and Dimensions, JEE Main
If z and ω are two complex numbers such that zω=1 and argz-arg(ω)=π2, then:

Options:
A zω¯=1-i2
B z¯ω=i
C zω¯=-1+i2
D z¯ω=-i

The correct answer is:
z¯ω=-i

Solution:
Let z=r and argz=θ, given argz-argω=π2

⇒θ-argω=π2

⇒argω=θ-π2

We know that, if z=r & argz=θ then 1z=1r & argz¯=-argz=-θ,

Hence, ω=1r and arg(ω¯)=-argω=π2-θ

Now, by using Euler form i.e. if z=r & argz=θ, then z=reiθ, we have z=reiθ and ω=1rei(θ-π2)

Also, z¯ω=r e-iθ×1r eiθ-π2

⇒z¯ω=e-iπ2

Using polar form i.e. eiθ=cosθ+isinθ, we get

z¯ω=1cos-π2+isin-π2=-i

⇒z¯ω=-i.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (10 Apr Shift 2)
If z and ω are two complex numbers such that zω=1 and argz-argω=3π2, then arg1-2z¯ω1+3z¯ω is:
(Here argz denotes the principal argument of complex number z)

Options:
A π4
B -3π4
C -π4
D 3π4

The correct answer is:
-3π4

Solution:
As

zω=1

If z=r, then ω=1r

Let argz=θ

∴ argω=θ-3π2

So,

z=reiθ

⇒z¯=rei-θ
⇒ω=1reiθ-3π2

Now, consider
1-2z¯ω1+3z¯ω=1-2ei-3π21+3ei-3π2=1-2i1+3i

=1-2i1-3i1+3i1-3i=1-5i+6i210=-5+5i10=-121+i

Then,

principal arg1-2z¯ω1+3z¯ω

=principal arg-121+i

=-π-π4=-3π4

Topic: Mathematics, Complex Number, JEE Main
If z and ω are two complex numbers such that zω=1 and argz-argω=3π2, then arg1-2z¯ω1+3z¯ω is:
(Here argz denotes the principal argument of complex number z)

Options:
A π4
B -3π4
C -π4
D 3π4

The correct answer is:
-3π4

Solution:
As

zω=1

If z=r, then ω=1r

Let argz=θ

∴ argω=θ-3π2

So,

z=reiθ

⇒z¯=rei-θ
⇒ω=1reiθ-3π2

Now, consider
1-2z¯ω1+3z¯ω=1-2ei-3π21+3ei-3π2=1-2i1+3i

=1-2i1-3i1+3i1-3i=1-5i+6i210=-5+5i10=-121+i

Then,

principal arg1-2z¯ω1+3z¯ω

=principal arg-121+i

=-π-π4=-3π4

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (20 Jul Shift 1)
If $\mathrm{z}$ and $\omega$ are two non-zero complex numbers such that $|z \omega|=1$ and $\operatorname{Arg}(z)-\operatorname{Arg}(\omega)=\frac{\pi}{2}$, then $\bar{z} \omega$ is equal to

Options:
A
$-\mathrm{i}$

B
1

C
$-1$

D
i.

The correct answer is:
$-\mathrm{i}$

Solution:
$|z \omega|=1$
$$
\text { As, } \operatorname{Arg}\left(\frac{z}{\omega}\right)=\frac{\pi}{2} \text { therefore } \frac{z}{\omega}=i
$$
$\therefore\left|\frac{z}{\omega}\right|=1$
From (1) \& (2), $|z|=|\omega|=1$ and $\frac{z}{\omega}+\frac{\bar{z}}{\omega}=0 ; z \bar{\omega}+\bar{z} \omega=0$
$\bar{z} \omega=-z \bar{\omega}=\frac{-z}{\omega} \cdot \bar{\omega} \cdot \omega ; \bar{z} \omega=-i|\omega|^2=-\mathrm{i}$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2003
If z and w are complex numbers satisfying zÂ¯+iwÂ¯=0 and ampzw=Ï, then ampw is equal to (where, ampwâ-Ï,Ï)

Options:
A π4
B -π4
C π2
D 3π4

The correct answer is:
π4

Solution:
Given z-+iw-=0⇒z-=-iw- or z=iw or zw=i
ampz-ampw=amp i=π2....i
also ampzw=π
ampz+ampw=π…..ii
Adding i & ii, we get,

2ampz=3π2
⇒ampz=3π4
Also, ampw=π4

Topic: Mathematics, Complex Number, JEE Main
If $z$ and $w$ are complex numbers such that $\bar{z}-i \bar{w}=0$ and $\operatorname{Arg}(z w)=\frac{3 \pi}{4}$, then $\operatorname{Arg} z=$

Options:
A $\frac{\pi}{16}$
B $\frac{\pi}{8}$
C $\frac{\pi}{4}$
D $\frac{3 \pi}{4}$

The correct answer is:
$\frac{\pi}{8}$

Solution:
We have,
$$
\begin{aligned}
\Rightarrow & i \bar{\omega} & =\bar{z} \Rightarrow \omega=\frac{1}{i} \bar{z} \\
\Rightarrow & \omega & =-\frac{1}{i} z \Rightarrow \omega=i z
\end{aligned}
$$
Now, we have
$$
\begin{aligned}
& \arg (z \omega)=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg (z(i z))=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg \left(i z^2\right)=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg (i)+\arg \left(z^2\right)=\frac{3 \pi}{4} \\
& {\left[\because \arg \left(z_1 z_2\right)=\arg \left(z_1\right)+\arg \left(z_2\right)\right]} \\
& \Rightarrow \quad \arg (i)+2 \arg (z)=\frac{3 \pi}{4} \quad\left[\because \arg \left(z^n\right)=n \arg (z)\right] \\
& \Rightarrow \quad \frac{\pi}{2}+2 \arg (z)=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg (z)=\frac{\pi}{8} \\
&
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main
If z and w are two non-zero complex numbers such that zw=1 and argz-argw=π2, then the value of 5iz-w is equal to

Options:
A –5
B 5i
C 5
D –5i

The correct answer is:
5

Solution:
argzw=Ï2⇒zw=zweiÏ2
⇒zw=zwi
⇒w=zwz-i
⇒wz-=z-zwz-i
⇒5iz-w=5-i2z2wz
=51zw
=5

Topic: Mathematics, Complex Number, JEE Main
If z be a complex number satisfying |z â4 + 8i|=4, then the least and the greatest value of |z + 2| are respectively (where i=-1 )

Options:
A 7 and 16
B 8 and 17
C 6 and 14
D 5 and 13

The correct answer is:
6 and 14

Solution:
|z+2|=|z+2–4+8i+4–8i|
|z+2|=|z–4+8i+6–8i|
z+2≤|z-4+8i|+|6-8i|
≤4+10=14
|z+2|=|z+2–4+8i+4–8i|
≥|z-4+8i|-|6-8i|
≥4-10=6

Topic: Mathematics, Complex Number, JEE Main
If Z=cosâ¡Ï+isinâ¡Ï âÏâÏ3,Ï, then the value of argZ2-Z is equal to (where, arg(Z) represents the argument of the complex number Z lying in the interval -Ï,Ï and i2=-1)

Options:
A 3ϕ+π2
B 3ϕ2
C 32ϕ-π
D 3ϕ-π2

The correct answer is:
32ϕ-π

Solution:
Z2-Z=ZZ-1
=cosâ¡Ï+isinâ¡Ïcosâ¡Ï-1+isinâ¡Ï
=cosâ¡Ï+isinâ¡Ï-2sin2Ï2+i2sinÏ2cosÏ2
=2isinÏ2cos3Ï2+isin3Ï2
=2sinÏ2cos3Ï+Ï2+isin3Ï+Ï2
Now, 3ÏâÏ,3Ïâ3Ï+Ï2âÏ,2Ï
But, the argument lies in -Ï,Ï, hence
argZ2-Z=3Ï+Ï2-2Ï=32Ï-Ï

Topic: Mathematics, Complex Number, JEE Main
If zâ i be any complex number such that z-iz+i is a purely imaginary number, then, z+1z is

Options:
A any non-zero real number other than 1 .
B a purely imaginary number.
C 0
D any non-zero real number

The correct answer is:
any non-zero real number

Solution:
Let z=x+iy
z-iz+i is a purely imaginary number
⇒x+iy-1x+iy+1×x-iy+1x-iy+1 is a purely imaginary
⇒x2+y2-1-i2xx2+y+12 is purely imaginary
⇒x2+y2-1=0⇒x2+y2=1…i
z+1z=x+iy+1x+iy
=x+iy+1x+iy×x-iyx-iy
=x+iy+x-iyx2+y2=2x
y≠±1 so x≠0 (from i and since, z won’t be an imaginary number)

Topic: Mathematics, Complex Number, JEE Main
If zi (where i=1,2,………..6) be the roots of the equation z6+z4-2=0, then ∑i=16zi4 is equal to

Options:
A 4
B 6
C 8
D 10

The correct answer is:
10

Solution:
Let, z2=t⇒t3+t2-2=0
⇒t-1t2+2t+2=0
⇒t=1 or t2+2t+2=0
⇒t=1,-1±i
⇒z=±1 or z2=-1±i
⇒z2=1,2⇒z4=1,2
⇒∑i=16zi4=1+1+2+2+2+2=10

Topic: Mathematics, Complex Number, JEE Main
If z is a complex number of unit modulus and argument θ, then arg 1+z1+z- can be equal to given z≠-1

Options:
A Î¸
B π - θ
C - θ
D π 2 - θ

The correct answer is:
θ

Solution:
Let z=cos θ+i sin θ
Then, 1+z1+z¯=1+cos θ+i sin θ1+cos θ-i sin θ
=2 cos2θ2+i·2 sinθ2cosθ22 cos2θ2-i·2 sinθ2cosθ2
=cosθ2+i sinθ2cosθ2-i sinθ2
=cosθ2+i sinθ22
=cos θ+i sin θ
∴ arg1+z1+z¯=arg z
=θ

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2013 (07 Apr)
If z is a complex number satisfying Rez+Imz=4, then z cannot be

Options:
A 172
B 10
C 7
D 8

The correct answer is:
7

Solution:
Let z=x+iy
i.e. x+y=4

Using perpendicular distance of line from origin, minimum value of z=22
Maximum value of z=4
So, z can’t be 7

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (09 Jan Shift 2)
If z is a complex number satisfying Rez+Imz=4, then z cannot be

Options:
A 172
B 10
C 7
D 8

The correct answer is:
7

Solution:
z=x+iy
x+y=4
Minimum value of z=22
Maximum value of z=4
So, z canât be 7

Topic: Mathematics, Complex Number, JEE Main
If z is a complex number satisfying the equation z-1+i2=2 and Ï=2z zâ 0, then the locus of Ï is

Options:
A x-y-1=0
B x+y-1=0
C x-y+z=0
D x+2y+1=0

The correct answer is:
x-y-1=0

Solution:
We have, z-1+i2=2
âx-12+y-12=2 (Putting z=x+iy )
âx2+y2=2x+yâ¦â¦..i
Let, Ï=h+ik=2z=2x+iy=2x-iyx2+y2, so
h=2xx2+y2,k=-2yx2+y2
âh-k=2x+yx2+y2=1 (from equation i )
â´ Locus of the point Ïh,k will be x-y=1.

Topic: Mathematics, Complex Number, JEE Main
If z is a complex number satisfying z2+3z2=-4, then the sum of the absolute values of the imaginary parts of the roots of the equation is

Options:
A 0
B 23
C 2
D 23+2

The correct answer is:
23+2

Solution:
z4+4z2+3=0
z2+1z2+3=0
⇒z2=-1 or -3
⇒z=±i,±3i
Hence, the required sum is 1+-1+3+-3

Topic: Mathematics, Complex Number, JEE Main
If z is a complex number such that z≤1, then the minimum value of z+123+4i is:

Options:
A 2
B 52
C 32
D 3

The correct answer is:
32

Solution:
Given: z+3+4i2; z≤1

Let, z=x+iy

And z=1 gives x2+y2=1, which represents a circle having centre at 0,0 and radius as 1 unit.

Now, z+3+4i2 gives the distance of z from -32,-2

So, we need to find the minimum distance of AP.

⇒AP=OP-OA

⇒AP=-32-02+0+22-1

⇒AP=94+4-1

⇒AP=52-1

⇒AP=32

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2024 (01 Feb Shift 2)
If z is a complex number such that z=2, then the area (in sq. units) of the triangle whose vertices are given by z, -iz and iz-z is equal to

Options:

The correct answer is:
6

Solution:
Let z=x+iy
∴ points will be x,y, y,-x, -x-y,-y+x
Area of the triangle for these points as vertices=32x2+y2 sq. units
∵z=x2+y2=2⇒x2+y2=4
∴ Area of triangle =6 sq. units

Topic: Mathematics, Complex Number, JEE Main
If z is a complex number such that z≥2, then the minimum value of z+12 :

Options:
A Is strictly greater than 5 2
B Is strictly greater than 3 2 but less than 5 2
C Is equal to 5 2
D Lies in the interval (1, 2)

The correct answer is:
Lies in the interval (1, 2)

Solution:
z≥2 i.e., z-0≥2
∴ The distance of 'z' from origin is greater than two.
We have to find minimum of z+12
z+12=z--12 = distance of z from -12, 0.
On the argand plane,
z lies in the dotted regions I, II, III, IV.
∴ The minimum distance is from the point B=2-12 = 3 2
⇒32∈1, 2

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2014 (06 Apr)
If z is a complex number such that z-iz-1 is purely imaginary, then the minimum value of |z-(3+3 i)| is :

Options:
A 32
B 22
C 22-1
D 62

The correct answer is:
22

Solution:
∵z-iz-1 is purely imaginary.

argz-iz-1=π2+nπ;n∈I

Geometrically, the locus of z is the circle with end points of diameter is 1,0 & 0,1 and centre is 12,12.So the nearest point to 3,3 on the circle is A1,1.

Minimum distance is AP=OP-OA

=32+32-12+12

=32-2

=22

Topic: Mathematics, Complex Number, JEE Main
If z is a complex number such that z-iz-1 is purely imaginary, then the minimum value of |z-(3+3 i)| is :

Options:
A 32
B 22
C 22-1
D 62

The correct answer is:
22

Solution:
∵z-iz-1 is purely imaginary.

argz-iz-1=π2+nπ;n∈I

Geometrically, the locus of z is the circle with end points of diameter is 1,0 & 0,1 and centre is 12,12.So the nearest point to 3,3 on the circle is A1,1.

Minimum distance is AP=OP-OA

=32+32-12+12

=32-2

=22

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (31 Aug Shift 2)
If z is a complex number, then the area of the triangle (in sq. units) whose vertices are the roots of the equation z3+iz2+2i=0 is equal to (where, i2=-1)

Options:

The correct answer is:
2

Solution:
z-iz2+2iz-2=0
⇒z=i,1-i,-1+i
Area=120111-11-1-11=-2=2 sq. units

Topic: Mathematics, Complex Number, JEE Main
If z is a complex number, then the number of common roots of the equation z1985+z100+1=0 and z3+2z2+2z+1=0, is equal to :

Options:
A 1
B 2
C 0
D 3

The correct answer is:
2

Solution:
Given,

z1985+z100+1=0 & z3+2z2+2z+1=0

Now solving,

z3+2z2+2z+1=0

⇒z+1z2-z+1+2zz+1=0

⇒z+1z2+z+1=0

⇒z=-1, z=ω, ω2

Now putting z=ω in z1985+z100+1 we get,

⇒ω1985+ω100+1

⇒ω2+ω+1=0 as ω3n=1

Also, z=ω2

⇒ω3970+ω200+1

⇒ω+ω2+1=0

Also, z=-1 will not satisfy the equation z1985+z100+1

Hence, there are two common root

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2024 (30 Jan Shift 2)
If z is a non-real complex number, then the minimum value of Im z5Im z5 is (Im z = Imaginary part of z)

Options:
A -2
B -4
C -5
D -1

The correct answer is:
-4

Solution:
Let, z=r(cosÎ¸+isinÎ¸)
z 5 = r 5 (cos5Î¸+isin5Î¸)
Im (z5)=r5sinâ¡5Î¸
and Im z5=r5sin5â¡Î¸
Im (z5)Im z5= sinâ¡5Î¸sin5â¡Î¸= A (Let)
âdAdÎ¸=5sin5â¡Î¸cosâ¡5Î¸-5sinâ¡5Î¸sin4â¡Î¸cosâ¡Î¸sin5â¡Î¸2
âdAdÎ¸=5sin4â¡Î¸sinâ¡Î¸cosâ¡5Î¸-sinâ¡5Î¸cosâ¡Î¸sin10â¡Î¸
â dA dÎ¸ = 5 sin 6 Î¸ sin â4Î¸ =0
âÎ¸=Ï4
The minimum value of A will be at Î¸=Ï4.
âsinâ¡5Ï4sinâ¡Ï45
=-12125
=-24=-4

Topic: Mathematics, Complex Number, JEE Main