StudentAidUnit 1: Unit 1 — Exercise 1 1
11th Class Mathematics · Unit 1: Complex Numbers
1.Find the multiplicative inverse of each of the following numbers: (i) (-4, 7) (ii) (\sqrt{2}, -\sqrt{5}) (iii) (1, 0)
Formula
(i) Using
(a,b)-1 = left(aa2+b2, -ba2+b2right)
Working
Putting the values
(-4,7)-1 = left(-4(-4)2+(7)2, -7(-4)2+(7)2right)
= left(-416+49, -716+49right)
Result
= left(-465, -765right)
Working
(ii)
(sqrt{2},-sqrt{5})-1 = left(frac{sqrt{2}{(sqrt{2})2+(-sqrt{5})2}, frac{sqrt{5}{(sqrt{2})2+(-sqrt{5})2}right)
Result
= left(frac{sqrt{2}{7}, frac{sqrt{5}{7}right)
Working
(iii)
(1,0)-1 = left(112+02, 012+02right)
Result
= (1,0)
2.Separate into real and imaginary parts (write as a simple complex number): (i) \frac{2-7i}{4+5i} (ii) \frac{(-2+3i)^2}{1+i} (iii) \frac{i}{1+i} (iv) \frac{(4+3i)^2}{4-3i}
(i) Multiply and divide by
4-5i
Working
2-7i4+5i × 4-5i4-5i = (2-7i)(4-5i)(4)2-(5i)2
= 8-10i-28i+35i216-25(-1)
= 8-38i+35(-1)16+25 = -27-38i41
Result
= -2741 - 3841i
(ii)
(-2+3i)21+i = (-2)2+(3i)2+2(-2)(3i)1+i
= 4+9i2-12i1+i = 4-9-12i1+i = -5-12i1+i
Multiply and divide by
1-i1-i
(-5-12i)(1-i)(1+i)(1-i) = -5+5i-12i+12i212-i2
= -5-7i+12(-1)1-(-1) = -5-7i-121+1
= -17-7i2 = -172-72i
(iii)
i1+i = i1+i × 1-i1-i = i(1-i)12-i2
= i-i21-(-1) = i-(-1)2 = 1+i2
= 12+12i
(iv)
= (4)2+(3i)2+2(4)(3i)4-3i
= 16+9i2+24i4-3i = 16-9+24i4-3i = 7+24i4-3i
Multiply and divide by
4+3i4+3i
(7+24i)(4+3i)(4-3i)(4+3i) = 28+21i+96i+72i2(4)2-(3i)2
= 28+117i-7216-9i2 = -44+117i16+9 = -44+117i25
= -4425+11725i
3.Prove that \bar z = z iff z is real.
Given
Let
z = x+iy quad dots (i)
bar z = x-iy
If
z=bar z ⇒ x+iy = x-iy ⇒ x+iy-x+iy=0 ⇒ 2iy=0
As
2ineq 0, y=0
Result
Put in equation (i)
z = x+i(0) ⇒ z=x, hence z is a real number
Conversely
suppose z is a real number; let z=x, z=x+0i, bar z = x-0i=x
Hence
bar z = z
4.For z \in C, show that: (i) \frac{z+\bar z}{2} = \operatorname{Re}(z) (ii) \frac{z-\bar z}{2i} = \operatorname{Im}(z) (iii) |z|^2 = z.\bar z
Given
(i) Let
z=a+bi, a,bin R
Then
bar z = overline{a+ib} ⇒ bar z = a-ib
Working
Now
z+bar z2 = a+bi+a-bi2
= 2a2 = a
Result
= Re(z)
Given
(ii) Let
z=a+ib
Then
bar z = a-ib
Working
Now
z-bar z2i = (a+ib)-(a-ib)2i
= a+ib-a+ib2i = 2ib2i
Result
= b = Im(z)
Given
(iii) Let
z=a+ib
Then
|z|=sqrt{a2+b2} ⇒ |z|2 = a2+b2 quad (i)
Now
bar z = a-ib,quad z.bar z = (a+ib)(a-ib)
= a2-(ib)2 = a2+b2 quad (ii)
Result
From (i) and (ii)
|z|2 = z.bar z
5.If z_1=2+i, z_2=3-2i, z_3=1+3i then express \frac{\bar z_1 \bar z_3}{z_2} in the form of a+ib.
Working
bar z1 bar z3z2 = (2+i)(1+3i)3-2i...
= (2-i)(1-3i)3-2i
= (2-3)+(-6-1)i3-2i = -1-7i3-2i
Multiply and divide by
(3+2i)
= (-1-7i)(3+2i)(3-2i)(3+2i)
Result
= (-3+14)+(-2-21)i32+22 = 1113-2313i
6.If z_1=2+7i and z_2=-5+3i, then evaluate the following: (i) |2z_1-4z_2| (ii) |3z_1+2\bar z_1| (iii) |-7z_2+2\bar z_2| (iv) |(z_1+z_2)^3|
(i) Firstly
2z1-4z2 = 2(2+7i)-4(-5+3i)
= 4+14i+20-12i = 24+2i
Now
|2z1-4z2| = sqrt{(24)2+(2)2}
Result
= sqrt{576+4} = sqrt{580} = 2sqrt{145}
(ii) Firstly
3z1+2bar z1 = 3(2+7i)+2(2-7i)
= 6+21i+4-14i = 10+7i
Now
|3z1+2bar z1| = sqrt{(10)2+(7)2}
= sqrt{100+49} = sqrt{149}
(iii) Firstly
-7z2+2bar z2 = -7(-5+3i)+2(-5-3i)
= 35-21i-10-6i = 25-27i
Now
|-7z2+2bar z2| = sqrt{(25)2+(-27)2}
= sqrt{625+729} = sqrt{1354}
(iv) Firstly
(z1+z2)3 = (2+7i-5+3i)3 = (-3+10i)3
= (-3)3+(10i)3+3(-3)2(10i)+3(-3)(10i)2
= -27+1000i3+270i-9(100i2)
= -27-1000i+270i+900
= 873-730i
Now
|(z1+z2)3| = sqrt{(873)2+(-730)2}
= sqrt{762129+532900} = sqrt{1295029} = 109sqrt{109}
7.Show that: i^{n+1}+i^{n+2}+i^{n+3}+i^{n+4}=0, for all n \in N.
Given
L.H.S
= in+1+in+2+in+3+in+4
= in+1(1+i+i2+i3)
= in+1(1+i-1-i) quad (As i2=-1 and i3=-i)
Result
= in+1(0) = 0 = R.H.S
8.Find the least positive value of n, if \left(\frac{1+i}{1-i}\right)^{2n}=1
1+i1-i = 1+i1-i × 1+i1+i = (1+i)212-i2
= 12+i2+2i1+1 = 1-1+2i2 = 2i2 = i
Given
So
1+i1-i=i
Working
Now
left(1+i1-iright)2n=1 ⇒ i2n=1 ⇒ i2n=i4 quad (As i4=1)
Result
By comparing
2n=4 ⇒ n=2
9.Show that, the value of i^n for n \in N and n>4 is i^r, where r is the remainder when n is divided by 4.
Given
By division algorithm
n=4q+r, where nin N and n>4, r is remainder
⇒ in = i4q+r = i4q.ir
= (i4)q.ir = (1)q.ir quad (As i4=1)
Result
in = ir quad Hence proved