1.Algebra Booster

4.56 Algebra Booster
19. Let z = r(cos q + i sin q)
Also, |z – 1| = 1
fi |r(cos q + i sin q) – 1| = 1
fi |(r cos q – 1) + i(r sin q)| = 1
2 2 2
fi ( r cos q – 1) + r sin q = 1
fi (r cos q – 1) 2 + r 2 sin 2 q = 1
fi r 2 – 2r cos q = 0
fi r = 2 cos q
z - 2 r(cosq+ isin q) -2
Now, =
z r(cos q+
isin q)
2 cos q(cos q+ isin q) -2
=
2 cos q(cos q+
isin q)
cos q(cos q+ i sin q) -1
=
(2 cos q(cos q+
i sin q))
2
(2 cos q+ i 2 cos qsin q) -2
=
2 cos q(cos q+
i sin q)
- 1 + cos 2q+
isin 2q
=
[(1 + cos 2 q) + isin 2 q]
i2sin2q
=
2 + 2cos2q
isin 2q
=
1 + cos 2q
i 2sinqcosq
=
2
2cos q
= i tan q
= i tan (Arg z)
Hence, the result.
20. Let z = r 1
(cos q 1
+ i sin q 1
)
and w = r 2
(cos q 2
+ i sin q 2
)
where |z| = r 1
, |w| = r 2
,
q 1
= Arg(z), q 2
= Arg(w)
Now, |z – w| 2
= (r 1
cos q 1
– r 2
cos q 2
) 2 + (r 1
sin q 1
– r 2
sin q 2
) 2
2 2
= r1 + r2 -2rr
1 2cos( q1-q2)
2 2
= r1 + r2 - 2rr 12+ 2rr 12-2rr
12cos( q1-q2)
= (r 1
– r 2
) 2 + 2r 1
r 2
(1 – cos (q 1
– q 2
))
2 2Êq1-
q2ˆ
= ( r1- r2) + 2rr
1 2◊2sin
Á
Ë
˜
2 ¯
2 2Êq1-
q2ˆ
= ( r1- r2) + 4rr
1 2sin
Á
Ë
˜
2 ¯
2
2 Êq1-
q2ˆ
£ ( r1- r2) + 4rr
1 2Á Ë
˜
2 ¯
= (r 1
– r 2
) 2 + r 1
r 2
(q 1
– q 2
) 2
(r 1
– r 2
) 2 + (q 1
– q 2
) 2 ( r 1
, r 2
£ 1)
= (|z| – |w|) 2 + [Arg(z) – Arg(w)] 2
Hence, the result.
21.
X¢
C
O
Y
Y¢
P
q
M
Let P = (x, y) = Z
x
y
cos q= sin q=
2 6 2 6
fi 1 x 2 6 y
= =
5 2 6 5 2 6
fi
2 6 24
x= y =
5 5
Thus, the complex number is
z = x + iy
2 6 24
= +i
5 5
22. We have,
25
z - = 24
z
fi |z 2 – 25| = 24|z|
fi 24|z| = |z| 2 – 25| ≥ |z 2 | – 25
fi |z 2 | – 25 £ 24|z|
fi |z 2 | – 24|z| – 25 £ 0
fi |z| 2 – 24|z| – 25 £ 0
fi |z| 2 – 25|z| + |z| – 25 £ 0
fi (|z| – 25)(|z| + 1) £ 0
fi 1 £ |z| £ 25
Thus, the greatest value of |z| is 25.
23.
Y
X¢
O
Y¢
Q
P
Here, C = (3, 4), CP = 5, OQ = 12
Thus, the minimum value of
|z 1
– z 2
|
= PQ
= OQ – OP
= 12 – 10
= 2
C
X
X