Q: Express (2+3i)2 in the form x+iy.

(2+3i)2=4+9i2+12i=4−9+12i=−5+12i

Q: If z1=3+7i,z2=−2i,z3=4+6i, find (i) z1−3z2+z3, (ii) z1z22, (iii) 4z12+2z23+z1z3+z2z3

(i) z1−3z2+z3=3+7i−3(−2i)+4+6i=7+19i (ii) z1z22=(3+7i)(−2i)2=(3+7i)×(−4)=−12−28i (iii) 4z12+====2z23+z1z3+z2z34(3+7i)2+2(−2i)3+(3+7i)(4+6i)+(−2i)(4+6i)4(9+49i2+42i)−16i3+(12+18i+28i+42i2)−(8i+12i2)4(9−49+42i)+16i+(12+46i−42)−(8i−12)−178+222i

Q: Express −5−4i2+3i in the form x+iy

−5−4i2+3i=(−5)2+42(2+3i)(−5+4i) (on multiplying both numerator and denominator by the conjugate of −5−4i ) =411(−10+8i−15i+12i2)=411(−22−7i) =−4122−417i

Q: If z1=1−i and z2=−2+4i, evaluate Imz1z1z2.

z1z1z2∴Im(z1)=2.z1z2=(1+i)(1−i)(−2+4i)=(12+12)(1−i)(−2+4i)(1−i)=2(1−i)2(−2+4i)=2−2i(−2+4i)=−i(−2+4i)

Q: Find 5−6i2+3i+5+6i2−3i

Conjugate of the complex number 5−6i2+3i+5−6i2+3i=5+6i2−3i Therefore, the given expression =2Re(5−6i2+3i) =2Re(52+62(2+3i)(5+6i))=612(10−18)=61−16

Q: If z1=2−5i,z2=3+4i, find z1z2−z2z1

Note that z2z1 is the conjugate of z1z2. Hence z1z2−z2z1=2iIm(z1z2) =2iIm{(2−5i)(3−4i)}=2i×(−23)=−46i

Q: If z1=4i,z2=2−7i,z3=3+i, find z1z2z3+z1z2z3.

We have z1z2z3+z1z2z3=2Re(z1z2z3) =2Re{4i(2+7i)(3−i)}=2Re{4i(13+19i)}=−2×76=−152.\]

Question 1

Two plane mirrors are placed at an angle θ between them. A beam of light parallel to one of the mirrors falls on the other mirror and after two successive reflections emerges parallel to the latter. Find the angle between the mirrors.

Accepted Answer

360−2θ=180+θ
3θ=180
θ=60∘

Question 2

A ray is reflected from a plane mirror. Without changing the direction of the incident ray, the mirror is rotated through an angle θ. Through what angle does the reflected ray rotate?

Accepted Answer

If a mirror is rotated through an angle θ, if the incident ray is kept same, then the reflected ray will turns through an angle of 2θ.

Question 3

Express (2+3i)2 in the form x+iy.

Accepted Answer

(2+3i)2​=4+9i2+12i=4−9+12i=−5+12i​

Question 4

If z1​=3+7i,z2​=−2i,z3​=4+6i, find
(i) z1​−3z2​+z3​,
(ii) z1​z22​,
(iii) 4z12​+2z23​+z1​z3​+z2​z3​

Accepted Answer

(i) z1​−3z2​+z3​=3+7i−3(−2i)+4+6i=7+19i
(ii) z1​z22​=(3+7i)(−2i)2=(3+7i)×(−4)=−12−28i
(iii) 4z12​+====​2z23​+z1​z3​+z2​z3​4(3+7i)2+2(−2i)3+(3+7i)(4+6i)+(−2i)(4+6i)4(9+49i2+42i)−16i3+(12+18i+28i+42i2)−(8i+12i2)4(9−49+42i)+16i+(12+46i−42)−(8i−12)−178+222i​

Question 5

Find the value of 1+i2+i4+i6+i8+i10.

Accepted Answer

Since the above expression is a geometric series with first term 1, common ratio i2 and number of terms 6 , sum of the series =1−i21−(i2)6​=(1+1)1−(−1)6​=0

Question 6

Express −5−4i2+3i​ in the form x+iy

Accepted Answer

−5−4i2+3i​=(−5)2+42(2+3i)(−5+4i)​
(on multiplying both numerator and denominator by the conjugate of −5−4i )
=411​(−10+8i−15i+12i2)=411​(−22−7i)
=−4122​−417​i

Question 7

If z1​=1−i and z2​=−2+4i, evaluate Imz1​​z1​z2​​.

Accepted Answer

z1​z1​z2​​∴Im(z1​​)=2.z1​z2​​​=(1+i)(1−i)(−2+4i)​=(12+12)(1−i)(−2+4i)(1−i)​=2(1−i)2(−2+4i)​=2−2i(−2+4i)​=−i(−2+4i)​

Question 8

Find 5−6i2+3i​+5+6i2−3i​

Accepted Answer

Conjugate of the complex number 5−6i2+3i​+5−6i2+3i​=5+6i2−3i​
Therefore, the given expression =2Re(5−6i2+3i​)
=2Re(52+62(2+3i)(5+6i)​)​=612​(10−18)=61−16​​

Question 9

If z1​=2−5i,z2​=3+4i, find z1​z2​​−z2​z1​​

Accepted Answer

Note that z2​z1​​ is the conjugate of z1​z2​​.
Hence z1​z2​​−z2​z1​​=2iIm(z1​z2​​)
=2iIm{(2−5i)(3−4i)}=2i×(−23)=−46i​

Question 10

If z1​=4i,z2​=2−7i,z3​=3+i, find z1​z2​z3​+z1​z2​z3​.

Accepted Answer

We have z1​z2​z3​+z1​z2​z3​=2Re(z1​z2​z3​)
=2Re{4i(2+7i)(3−i)}=2Re{4i(13+19i)}=−2×76=−152.​\]

Question 11

If z1​=2i,z2​=3​+i,z3​=1+i2​, find z3​z1​z2​​−z3​z1​z2​​.

Accepted Answer

We have z3​z1​zˉ2​​−z3​zˉ1​z2​​=2iIm(z3​z1​zˉ2​​)=2iIm(1+i2​2i(3​−i)​)
=2iIm(1+22i(3​−i)(1−i2​)​)=2iIm(32i​[(3​−2​)−i(6​+1)])=2i×32(3​−2​)​=34(3​−2​)​i.​

Question 12

If a+ib=(3−4i)(2+5i), find a2+b2

Accepted Answer

a2+b2​=∣(a+ib)∣2=∣(3−4i)(2+5i)∣2=(32+42)(22+52)=25×29=725​

Question 13

:
Solve for x and y:(5x−4y)+(2x+3y)i=30−11i where, x and y are real.

Accepted Answer

Equating real and imaginary parts on both sides,
(5 x-4 y)=30 (2 x+3 y)=-11
Solving for x and y, we obtain x=2,y=−5.

Question 14

Solve for x and y:(5x−4y)+(2x+3y)i=30−11i where, x and y are real.

Accepted Answer

Equating real and imaginary parts on both sides,
(5 x-4 y)=30 (2 x+3 y)=-11
Solving for x and y, we obtain x=2,y=−5.

Question 15

Find the square roots of 7+24i

Accepted Answer

Let (a+ib) represent a square root of (7+24i)
We have (a+ib)2=7+24i
⇒⇒​a2−b2=7,2ab=24(a2+b2)2=(a2−b2)2+4a2 b2=625 or a2+b2=25​
From the two equations a2−b2=7,a2+b2=25, we get a2=16, b2=9 ⇒a=±4, b=±3.
The two square roots are (4+3i) and (−4−3i).

Question 16

If (1+2i)(1+3i)(1+4i)…(1+ni)=x+iy, find the value of 5.10.17…(n2+1)

Accepted Answer

We have, ∣(1+2i)(1+3i)…(1+ni)∣2=∣x+iy∣2 ⇒5⋅10.17…(n2+1)=x2+y2.

Question 17

If x=2+3i, compute the value of 3x4−14x3+48x2−30x+25.

Accepted Answer

We have, x−2=3i
Squaring both sides, x2−4x+4=−9 or x2−4x+13=0
3x4−14x3+48x2−30x+25=3x2(x2−4x+13)−2x(x2−4x+13)+(x2−4x+13)+12=12, by (1)​

Question 18

If the number z+2z−2​ is pure imaginary, prove that ∣z∣=2.

Accepted Answer

Given: z+2z−2​ is pure imaginary ⇒arg(z+2z−2​)=2π​
Now, arg(z+2z−2​) represents the angle between PA and PB where P represents z,A represents (1,0) and B represents (−1,0). Since angle APB =2π​, P must be on the circle on AB as diameter. Clearly, the centre of the circle is at the origin and its radius is equal to 2 .
Remark
If the number z+2z−2​ is to be real, we note that z has to be on the x-axis.

Question 19

Prove that (23​+i​)6+(2i−3​​)6=−2

Accepted Answer

23​​+2i​=cos6π​+isin6π​(23​​+2i​)6=cosπ+isinπ=−1(23​​−2i​)6=cosπ−isinπ=−1​
Result follows.

Question 20

a,b,c are positive and a<b<c. If the roots of the quadratic equation ax2+bx+c=0 are complex, prove that their modulii are greater than 1 .

Accepted Answer

ax2+bx+c=0
The roots are 2a−b±b2−4ac​​=2a−b±i4ac−b2​​, since b2−4ac<0. ∣∣​2a−b±i4ac−b2​​∣∣​=4a2b2+4ac−b2​=ac​>1

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Solutions for IIT-JEE Super Course in Mathematics - Algebra II by Trishna Knowledge Systems

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