StudentAidUnit 1: Complex Numbers — Exercise 1 3
10th Class Mathematics · Unit 1: Complex Numbers
1.3.1.Find the modulus of the following complex numbers: (i) 4+3i (ii) -5-4i (iii) 3/5 - 4/5 i (iv) -√2 - √3 i
Given
(i)
z=4+3i
Formula
|z|=sqrt{a2+b2}
=sqrt{42+32}=sqrt{25}
Result
(i)
|z|=5
Given
(ii)
z=-5-4i
|z|=sqrt{(-5)2+(-4)2}=sqrt{41}
Result
(ii)
|z|=sqrt{41}
Given
(iii)
z=frac35-frac45 i
|z|=sqrt{left(frac35right)2+left(-frac45right)2}=sqrt{925+1625=sqrt{2525
Result
(iii)
|z|=1
Given
(iv)
z=-sqrt2-sqrt3 i
|z|=sqrt{(-sqrt2)2+(-sqrt3)2}=sqrt{2+3}
Result
(iv)
|z|=sqrt5
1.3.2.If z1 = 2+7i and z2 = 4-3i, then verify that: (i) conj(z1+z2) = conj(z1)+conj(z2) (ii) conj(z1 z2) = conj(z1) conj(z2) (iii) conj(z1/z2) = conj(z1)/conj(z2)
Given
(i)
overline{z1+z2} = overline{z1}+overline{z2}
L.H.S = overline{z1}+overline{z2} = (2-7i)+(4+3i) = 6-4i
R.H.S = overline{z1+z2} = overline{(2+7i)+(4-3i)} = overline{6+4i} = 6-4i
Result
(i)
L.H.S=R.H.S=6-4i
Given
(ii)
overline{z1z2} = overline{z1} overline{z2}
z1z2 = (2+7i)(4-3i) = 8+22i+21 = 29+22i ⇒ overline{z1z2}=29-22i
overline{z1} overline{z2} = (2-7i)(4+3i) = 8-22i+21 = 29-22i
Result
(ii)
L.H.S=R.H.S=29-22i
Given
(iii)
overline{left(z1z2right)} = frac{overline{z1}{overline{z2}
z1z2 = 2+7i4-3i × 4+3i4+3i = 8+34i-2125 = -13+34i25
This is z1/z2 itself; the book labels it ‘L.H.S’ but never takes its conjugate (as it correctly does in part (ii)), so the printed L.H.S value -13+34i over 25 does not actually equal the R.H.S value computed below.
frac{overline{z1}{overline{z2} = 2-7i4+3i × 4-3i4-3i = 8-34i-2125 = -13-34i25
Result
(iii)
L.H.S=R.H.S=-13-34i25
1.3.3.If z = 5-2i, then verify that: (i) conj(conj(z)) = z (ii) |z| = |conj(z)| (iii) |z| = |-z| (iv) z·conj(z) = |z|^2 (v) |z| = |-conj(z)|
Given
(i)
overline{overline{z} = z, quad z=5-2i
overline{z}=5+2i ⇒ overline{overline{z} = overline{5+2i} = 5-2i = z
Result
(i)
Verified
Given
(ii)
|z| = |overline{z}|
|z|=sqrt{52+(-2)2}=sqrt{29}; quad |overline z|=sqrt{52+22}=sqrt{29}
Result
(ii)
|z|=|overline z|=sqrt{29}
Given
(iii)
|z| = |-z|
-z=-5+2i; quad |-z|=sqrt{(-5)2+22}=sqrt{29}
Result
(iii)
|z|=|-z|=sqrt{29}
Given
(iv)
zoverline{z} = |z|2
zoverline z = (5-2i)(5+2i) = 25-4i2 = 25+4 = 29; quad |z|2=(sqrt{29})2=29
Result
(iv)
zoverline z = |z|2 = 29
Given
(v)
|z| = |-overline{z}|
-overline z = -5-2i; quad |-overline z| = sqrt{(-5)2+(-2)2}=sqrt{29}
Result
(v)
|z|=|-overline z|=sqrt{29}
1.3.4.If z = 4-3i, then verify that: (i) |z| = |-z| (ii) |z| = |conj(z)| (iii) |z| = |-conj(z)| (iv) z·conj(z) = |z|^2 (v) |z| = |-conj(z)|
Given
(i)
|z|=|-z|, quad z=4-3i
|z|=sqrt{42+(-3)2}=sqrt{25}=5; quad -z=-4+3i, |-z|=sqrt{16+9}=5
Result
(i)
|z|=|-z|=5
Given
(ii)
|z|=|overline z|
overline z=4+3i; quad |overline z|=sqrt{16+9}=5
Result
(ii)
|z|=|overline z|=5
Given
(iii)
|z|=|-overline z|
-overline z=-4-3i; quad |-overline z|=sqrt{16+9}=5
Result
(iii)
|z|=|-overline z|=5
Given
(iv)
zoverline z = |z|2
zoverline z = (4-3i)(4+3i) = 16-9i2 = 25; quad |z|2=52=25
Result
(iv)
zoverline z = |z|2 = 25
Given
(v)
|z|=|-overline z|
From (iii), |-overline z|=5; also |z|=5
Result
(v)
|z|=|-overline z|=5
1.3.5.If z1 = 2+3i, z2 = -1+i, then evaluate: (i) Re(z1 z2) (ii) Im(z1 z2) (iii) |z1| (iv) |z2| (v) |z1+z2| (vi) |z1-z2|
Given
(i)
Re(z1z2)
z1z2 = (2+3i)(-1+i) = -2+2i-3i+3i2 = -5-i
Result
(i)
Re(z1z2) = -5
Given
(ii)
Im(z1z2)
From above, z1z2 = -5-i
Result
(ii)
Im(z1z2) = -1
Given
(iii)
|z1|
|z1| = sqrt{22+32} = sqrt{13}
Result
(iii)
|z1| = sqrt{13}
Given
(iv)
|z2|
|z2| = sqrt{(-1)2+12} = sqrt{2}
Result
(iv)
|z2| = sqrt2
Given
(v)
|z1+z2|
z1+z2 = 1+4i; quad |z1+z2| = sqrt{12+42} = sqrt{17}
Result
(v)
|z1+z2| = sqrt{17}
Given
(vi)
|z1-z2|
z1-z2 = 3+2i; quad |z1-z2| = sqrt{32+22} = sqrt{13}
Result
(vi)
|z1-z2| = sqrt{13}