Page 165 - Algebra

Explanation:
7i4 – 3i7
1.
2. i20+i0
=(i2)10 +1=(–1)10 +1=1+1=2
3. = 6+5𝑖 2−3𝑖
4.
22 23 = 7(i ) – 3(i ) .i
= 7 + 3i
Multiply and divide by 2 + 3i
= 6+5𝑖 × 2+3𝑖 2−3𝑖 2+3𝑖
= 12+18𝑖+10𝑖+15𝑖2 = 22−(3𝑖)2
5.
6.
= 28𝑖−3 or 13
(p – a) + (–q – b)i = 3p + 0i
12+28𝑖−15
−3 + 28𝑖 13 13
4+9
Simplify
= a + bi
−3 28 So, a = 13 and b = 13
(p – qi) + (– a – bi) = 3p
p – a – qi – bi = 3p
p – a = 3p and –q – b = 0
a = –2p and b = –q
Put the values in (p + qi) (a – 2bi)
= (p + qi) (a – 2bi)
= (p + qi) (–2p + 2qi)
= –2p2 + 2pqi – 2pqi + 2q2i2
= –2p2 –2q2
= 2(–q2 – p2)
=2−𝑖 ×8−6𝑖 8+6𝑖 8−6𝑖
= 16−12𝑖−8𝑖+6𝑖2 82−(6𝑖)2
=
16−20𝑖−6
10−20𝑖 1 𝑖
==–
64+36
100 10 5
Put p(2i) = 2 in the function.
p(2i) = b
2=b
b2 = 4
b = 2 or –2
√
17 − (2𝑖)4
√17 − (2𝑖)4
Square both the sides
4=b2 (17–16)
Page 164 of 177
Algebra I & II