1.Algebra Booster

Complex Numbers 4.37
Hence, the solutions are
Ï Ê1 3ˆ Ê1 3ˆ¸
Ì1± 2 i, Á - i ˜,
Á + i ˜˝
Ó Ë2 2 ¯ Ë2 2 ¯˛
24. Thus, the greatest and the least value of |z| are
13 + 1, 13 - 1 respectively.
25. Thus, the greatest and the least value of |z| are
34 + 2, 34 - 2 .
Hence, the required difference
= ( 34 + 2) -( 34 -2)
= 4
26. Given a + ib = (1 + i)(1 + 2i)(1 + 3i) … (1 + ni)
fi |a + ib| = |(1 + i)(1 + 2i)(1 + 3i) … (1 + ni)|
= |(1 + i)||(1 + 2i)||(1 + 3i)| … |(1 + ni)|
fi
2 2 2 2 2 2 2 2 2 2
a + b = 1 + 1 1 + 2 1 + 3 … 1 + n
fi (a 2 + b 2 ) = 2.5.10 … (n 2 + 1)
Hence, the result.
27. We have,
a + ib
x+ iy =
a - ib
a + ib
fi | x+ iy|
=
a - ib
fi
| a + ib|
=
| a - ib|
fi
2 | a + ib|
| x+ iy|
=
2
| a - ib|
2 2
2 2 ( a + b )
fi x + y = = 1
2 2
( a + b )
Hence, the result.
28. We have,
fi
fi
z + |z| = 2 + 8i
2
2 2
( x+ iy) + x + y = 2+
8i
2 2
( x+ x + y ) + iy = 2+
8i
2 2
Thus, x + x + y = 2, y = 8
fi
fi
2
( x+ x + 64) = 2
2
x + 64 = (2 – x)
fi (x 2 + 64) = (2 – x) 2
fi = 4 – 4x + x 2
fi 4x = –60
fi x = –15
2 2
Hence, || z = x + y
= 225 + 64 = 289 = 17
29. We have,
i p i p
iz
iq
2 iq
( + q
2 )
= ire = e re = re
Ê Êp
ˆ Êp
ˆˆ
= rÁcos
+ q + isin
Á + q˜
Ë
Á
Ë
˜
2 ¯ Ë2
¯˜
¯
= r(–sin q + i cos q)
Now, e iz = e r(–sin q + i cos q) –r sin q + i(r cos q)
= e
= e –r sin q –i(r cos q)
e
fi |e iz | = |e –r sin q e –i(r cos q) |
= e –r sin q |e –i(r cos q) |
= e –r sin q ¥ 1
–r sin q
= e
30. We have,
ab + ba
| ab|
| ab + ba|
=
| ab|
| ab| + | ba|
£
| ab|
| ab || | + | ba || |
=
| ab || |
| ab || | | ba || |
= +
| ab || | | ab || |
= 2 ( | a| = | a|,| b| = | b|)
Thus, the maximum value is 2.
31. Given |z 1
| = 1
fi |z 1
| 2 = 1 fi z 1
◊z – = 1 1
fi
1
z 1 =
z1
1 1
Similarly, z2 = , z3
=
z z
Now,
1 2 3
2 3
1 1 1
+ + = 1
z z z
| z + z + z | = 1
fi 1 2 3
| z + z + z | = 1
fi 1 2 3
fi |z 1
+ z 2
+ z 3
| = 1
32. We have z = (3 + 7i)(p + iq), p, q ΠI Р{0}
fi z = (3p – 7q) + i(7p + 3q)
fi 2 2
|| z = (3 p –7) q + (7p+
3) q
fi |z| 2 = (3p – 7q) 2 + (7p + 3q) 2
= 9p 2 + 49q 2 + 49p 2 + 9q 2
= 58(p 2 + q 2 ) …(i)
Since z is purely imaginary number, so
(3p – 7q) = 0