Mathematics Test

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Mathematics Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

The complex number z=1+i represented by the point P in argand plane and OP is rotated by an angle of (π/2) in counter clock wise direction then the resulting complex number is:

A
\({z}\)

B
z

C
\(-\bar{z}\)

D
\(\frac{\bar{z}}{|z|^{2}}\)

Solution

\(z=1+i\)
\(\therefore \arg z=\tan ^{-1}\left(\frac{1}{1}\right)=\frac{\pi}{4}\)
Let the complex number obtained after rotation be \(z_{1}\) \(\therefore \arg \left(z_{1}\right)=\frac{\pi}{4}+\frac{\pi}{2}\)
\(=\frac{3 \pi}{4}\)
\(z_{1}=-1+i\)
Now, \(\bar{z}=1-i\)
Hence, \(z_{1}=-\bar{z}\)
Hence, the correct option is (C)
Question 2
1 / -0

If a>0 and z|z|+az+2a=0, then z must be-

A
purely imaginary

B
a positive real number

C
a negative real number

D
having positive imaginary part

Solution

\(z|z|+\boldsymbol{a} z+\mathbf{2} \boldsymbol{a}=(\boldsymbol{x}+i y) \sqrt{\boldsymbol{x}^{2}+\boldsymbol{y}^{2}}+\boldsymbol{a}(\boldsymbol{x}+i \boldsymbol{y})+2 \boldsymbol{a}=\mathbf{0}\)
by comparsion \(y \sqrt{x^{2}+y^{2}}+a y=0\)
\(y=0\)
\(x \sqrt{x^{2}+y^{2}}+a x+2 a=0\)
\(" \mathrm{a}^{\text {" }}\) is positive so \(\mathrm{x}\) should be negative.
\(\therefore z\) is a negative real number.
Hence, the correct option is (C)
Question 3
1 / -0

If 2i²+6i³+3i¹⁶−6i¹⁹+4i²⁵=x+iy, then (x,y)=

A
(1,4)

B
(4,1)

C
(−1,4)

D
(−1,−4)

Solution

The value of \(2 i^{2}+6 i^{3}+3 i^{16}-6 i^{19}+4 i^{25}\)
\(=-2-6 i+\left[3(-1)^{8}\right]-\left[6 i^{18} \cdot i\right]+\left[4 i^{24} \cdot i\right]\)
\(=-2-6 i+3-\left[6(-1)^{9} \cdot i\right]+\left[4(-1)^{12} \cdot i\right]\)
\(=1-6 i+6 i+4 i\)
\(=1+4 i\)
\(=x+i y\)
Hence, \((x, y)=(1,4)\)
Hence, the correct option is (A)
Question 4
1 / -0

The necessary and sufficient condition for the points z₁, z₂, z₃ to be collinear is that-

A
(z₃−z₂)/(z₂−z₁) is purely real

B
(z₂+z₃)/(z₂+z₁) is purely real

C
(z₃+z₁)/(z₃+z₂) is purely imaginary

D
(z₂−z₁)/(z₃−z₂) is purely imaginary

Solution

Let \(z_{1}=r_{1} e^{i \theta_{1}}, z_{3}=r_{3} e^{i \theta_{3}}\)
\& \(x_{2}\) be origin \(\frac{z_{3}-z_{2}}{z_{2}-z_{1}}\)
\(=\frac{r_{3} e^{i \theta_{3}}}{-r_{1} e^{i \theta_{1}}}\)
\(=\frac{r_{3}}{-r_{1}} e^{i\left(\theta_{1}-\theta_{1}\right)}\)
For the points to be collinear \(\left(\theta_{3}-\theta_{1}\right)=0\) \(\frac{z_{3}-z_{2}}{z_{2}-z_{1}}=\frac{r_{3}}{r_{1}}\)
and we know that \(\frac{r_{3}}{r_{1}}\) is real \(\frac{z_{3}-z_{2}}{z_{2}-z_{1}}\) is purely real.
Hence, the correct option is (A)
Question 5
1 / -0

z₁,z₂ are the roots of the equation z²+az+b=0. Let z₁, z₂ and the origin be the vertices of an equilateral triangle. Then a²−3b=

A
0

B
1

C
-1

D
2

Solution

For equilateral \(\Delta\), \(z_{1}^{2}+z_{2}^{2}+z_{3}^{2}-z_{1} z_{2}-z_{2} z_{3}-z_{3} z_{1}=0\)
We have \(z_{3}=0\)
\(\therefore z_{1}^{2}+z_{2}^{2}-z_{1} z_{2}=0\)
(1)
\(z_{1} \& z_{2}\) are roots of \(z^{2}+a z+b\)
\(\therefore z_{1}+z_{2}=-a\)
\(z_{1} z_{2}=b\)
\(\therefore z_{1}^{2}+z_{2}^{2}=a^{2}-2 b\)
purting the values in equation 1 \(a^{2}-2 b-b=0\)
\(\therefore a^{2}-3 b=0\)
Hence, the correct option is (A)
Question 6
1 / -0

If \(i^{2}=-1\) then the value of \(\sum_{n=1}^{200} i^{2 n}\) is:

A
50

B
10

C
0

D
100

Solution

As \(i^{2}=-1\)
\(\Rightarrow i^{2 n}=(-1)^{n}\)
For \(n=\) odd \(, i^{2 n}=-1\)
For \(n=\) even, \(i^{2 n}=1\)
There are equal number of even and odds in between 1 to 200
\(\Rightarrow \sum_{n=1}^{200}=-1+1-1+1-\ldots \ldots .=0\)
Hence, the correct option is (C)
Question 7
1 / -0

(1+i+i²+i³+i⁴+i⁵)(1+i)=

A
i

B
2i

C
3i

D
4i

Solution

\(\mathbf{1}+i+i^{2}+i^{3}+i^{4}+i^{5}\) forms a G.P series
with a common ratio of i'. Hence \(1+i+i^{2}+i^{3}+i^{4}+i^{5}\)
\(=\frac{1-i^{6}}{1-i}\)
\(=\frac{1-\left((i)^{2}\right)^{3}}{1-i}\)
\(=\frac{2}{1-i}\)
Now \(\left(1+i+i^{2}+i^{3}+i^{4}+i^{5}\right)(i+1)\)
\(=\left(\frac{2}{1-i}\right)(i+1)\)
\(=\frac{2(i+1)}{1-i}\)
\(=\frac{2(i+1)^{2}}{(1-i)(i+1)}\)
\(=\frac{2(1-1+2 i)}{2}\)
\(=2 i\)
Hence, the correct option is (B)
Question 8
1 / -0

If a,b,c are real and a+b+c=0 (for at least one of a,b,c non zero ) and az₁+bz₂+cz₃=0 then z_1,z₂,z₃are

A
Vertices of equilateral triangle

B
Vertices of an isosceles triangle

C
Vertices of right angled triangle

D
Collinear points

Solution

The necessary and sufficient condition for the points \(z_{1}, z_{2} \& z_{3}\) to be collinear is \(\frac{z_{1}-z_{3}}{z_{2}-z_{3}}\) ie purely real. Given that \(a z_{1}+b z_{2}+c z_{3}=0\)
\(a z_{1}+b z_{2}+(-a-b) z_{3}=0\)
\(a+b+c=0]\)
\(a\left(z_{1}-z_{3}\right)+b\left(z_{2}-z_{3}\right)=0\)
\(\frac{z_{1}-z_{3}}{z_{2}-z_{3}}=\frac{-b}{a} \quad[\therefore a, b, c\) are real \(]\)
\(\frac{-b}{a}\) is purely real
Therefore \(z_{1}, z_{2} \& z_{3}\) are collinear points.
Hence, the correct option is (D)
Question 9
1 / -0

If z₁, z₂, z_3,are in A.P, then which of the following is/are true?

A
The three points are collinear

B
z₁ is the mid point of z₂, z₃

C
z₃ is the mid point of z₁, z₂

D
The three points are the vertices of an equilateral triangle

Solution

Let \(z_{d}\) be the common difference between complex numbers, then, \(z_{2}=z_{1}+z_{d}\)
\(z_{3}=z_{1}+2 z_{d}\)
we see that, \(\left|z_{2}-z_{1}\right|=\left|z_{d}\right|\)
\(\left|z_{3}-z_{2}\right|=\left|z_{d}\right|\)
and \(\left|z_{3}-z_{1}\right|=2\left|z_{d}\right|\)
We find that the sum of distances of \(z_{2}\) from \(z_{3}\) and \(z_{1}\) equals distance between \(z_{3}\) and \(z_{1}\) which is possible only if. \(z_{1}, z_{2}, z_{3}\) lie on a straight line.
Hence, the correct option is (A)
Question 10
1 / -0

\(i^{57}+\frac{1}{i^{125}}=\)

A
0

B
2i

C
-2i

D
2

Solution

\(i^{57}+\frac{1}{i^{125}}\)
\(=\left(i^{56} \cdot i\right)+\frac{i}{i^{126}}\)
\(=\left(\left(i^{2}\right)^{28} \cdot i\right)+\frac{i}{\left(i^{2}\right)^{63}}\)
\(=\left((-1)^{28} \cdot i\right)+\frac{i}{(-1)^{63}}\)
\(=i-i\)
\(=0\)
Hence, the correct option is (A)

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Mathematics Test - 27

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Mathematics Test - 27

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Question 1

\({z}\)

z

\(-\bar{z}\)

\(\frac{\bar{z}}{|z|^{2}}\)

Solution

Question 2

purely imaginary

a positive real number

a negative real number

having positive imaginary part

Solution

Question 3

(1,4)

(4,1)

(−1,4)

(−1,−4)

Solution

Question 4

(z3−z2)/(z2−z1) is purely real

(z2+z3)/(z2+z1) is purely real

(z3+z1)/(z3+z2) is purely imaginary

(z2−z1)/(z3−z2) is purely imaginary

Solution

Question 5

0

1

-1

2

Solution

Question 6

50

10

0

100

Solution

Question 7

i

2i

3i

4i

Solution

Question 8

Vertices of equilateral triangle

Vertices of an isosceles triangle

Vertices of right angled triangle

Collinear points

Solution

Question 9

The three points are collinear

z1 is the mid point of z2, z3

z3 is the mid point of z1, z2

The three points are the vertices of an equilateral triangle

Solution

Question 10

0

2i

-2i

2

Solution

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(z₃−z₂)/(z₂−z₁) is purely real

(z₂+z₃)/(z₂+z₁) is purely real

(z₃+z₁)/(z₃+z₂) is purely imaginary

(z₂−z₁)/(z₃−z₂) is purely imaginary

z₁ is the mid point of z₂, z₃

z₃ is the mid point of z₁, z₂