StudentAidUnit 1: Complex Numbers — Exercise 1 4
10th Class Mathematics · Unit 1: Complex Numbers
1.4.1.Find the real and imaginary parts of the following complex numbers: (i) (8-3i)^2 (ii) (5+3i)^-1 (iii) (4-5i)^-1 (iv) (4-3i)^-2 (v) ((3+2i)/(4+3i))^-1 (vi) ((2-i)/(2+i))^-2 (vii) ((1-2i)/(1+i))^2
Given
(i)
(8-3i)2
= 64-48i+9i2 = 55-48i
Result
(i)
Real=55, Imaginary=-48
Given
(ii)
(5+3i)-1
= 15+3i × 5-3i5-3i = 5-3i34
Result
(ii)
Real=534, Imaginary=-334
Given
(iii)
(4-5i)-1
= 4+5i42+52 = 4+5i41
Result
(iii)
Real=441, Imaginary=541
Given
(iv)
(4-3i)-2
= left(4+3i25right)2 = 16+24i+9i2625 = 7+24i625
Result
(iv)
Real=7625, Imaginary=24625
Given
(v)
left(3+2i4+3iright)-1
= 4+3i3+2i × 3-2i3-2i = 12-8i+9i-6i29+4 = 18+i13
Result
(v)
Real=1813, Imaginary=113
Given
(vi)
left(2-i2+iright)-2
= left(2+i2-iright)2 = left((2+i)24+1right)2-style = (3+4i)225 = 9+24i-1625
Result
(vi)
Real=-725, Imaginary=2425
Given
(vii)
left(1-2i1+iright)2
1-2i1+i × 1-i1-i = -1-3i2 ⇒ left(-1-3i2right)2 = 1+6i-94 = -8+6i4
Result
(vii)
Real=-2, Imaginary=frac32
1.4.2.Solve the following simultaneous linear equations with complex coefficients for w and z: (i) 3z+(2+i)w=11-i, (2-i)z-w=-1+i (ii) 2z+(3+i)w=9-i, -iz-iw=-1+i (iii) z-4w=3i, 2z+3w=11-5i (iv) z+w=3i, 2z+3w=2 (v) 2z+(3+i)w=1, -z-(1-i)w=2
Given
(i)
3z+(2+i)w=11-i quad (1); quad (2-i)z-w=-1+i quad (2)
From (2): w=(2-i)z+1-i; substitute in (1): 3z+(2+i)[(2-i)z+1-i]=11-i
3z+5z+(3-i)=11-i ⇒ 8z=8 ⇒ z=1 ⇒ w=(2-i)(1)+1-i=3-2i
Result
(i)
w=3-2i, z=1
Given
(ii)
2z+(3+i)w=9-i quad (1); quad -iz-iw=-1+i quad (2)
From (2): z+w=1-ii=-1-i ⇒ z=-1-i-w
Substitute in (1): 2(-1-i-w)+(3+i)w=9-i ⇒ w(1+i)=11+i ⇒ w=11+i1+i=6-5i
z=-1-i-(6-5i)=-7+4i
Result
(ii)
w=6-5i, z=-7+4i
Given
(iii)
z-4w=3i quad (1); quad 2z+3w=11-5i quad (2)
From (1): z=3i+4w; substitute in (2): 2(3i+4w)+3w=11-5i ⇒ 11w=11-11i ⇒ w=1-i
z=3i+4(1-i)=4-i
Result
(iii)
w=1-i, z=4-i
Given
(iv)
z+w=3i quad (1); quad 2z+3w=2 quad (2)
From (1): z=3i-w; substitute in (2): 2(3i-w)+3w=2 ⇒ w=2-6i
z=3i-(2-6i)=-2+9i
Result
(iv)
w=2-6i, z=-2+9i
Given
(v)
2z+(3+i)w=1 quad (1); quad -z-(1-i)w=2 quad (2)
From (2): z=-2-(1-i)w; substitute in (1): -4+(1+3i)w=1 ⇒ w=51+3i=1-3i2
z=-2-(1-i) · 1-3i2=-2--2-4i2=-1+2i
Result
(v)
w=1-3i2, z=-1+2i