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KEAM 2012 Math Paper
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© examsnet.com
Question : 31
Total: 120
Let
z
1
=
2
√
3
+
i
6
√
7
6
√
7
+
i
2
√
3
and
z
2
=
√
11
+
i
3
√
13
3
√
13
−
i
√
11
.
Then
|
1
z
1
+
1
z
2
|
is equal to
47
264
|
z
1
−
z
2
|
|
z
1
+
z
2
|
|
z
1
z
2
|
Validate
Solution:
Given,
z
1
=
2
√
3
+
i
6
√
7
6
√
7
+
i
2
√
3
×
6
√
7
−
i
2
√
3
6
√
7
−
i
2
√
3
=
12
√
21
+
12
√
21
+
252
i
−
12
i
252
+
12
=
24
√
21
+
240
i
264
=
√
21
11
+
10
i
11
and
z
2
=
√
11
+
i
3
√
13
3
√
13
−
i
√
11
×
(
3
√
13
+
i
√
11
)
(
3
√
13
+
i
√
11
)
=
3
√
143
−
3
√
143
+
117
i
+
11
i
117
+
11
=
128
i
128
=
i
Now,
1
z
1
=
z
1
|
z
1
|
2
=
√
21
11
−
10
i
11
√
21
121
+
100
121
=
√
21
11
−
10
i
11
=
z
1
and
1
z
2
=
z
2
|
z
2
|
2
=
−
i
√
1
=
−
i
=
z
2
∴
1
n
1
+
1
z
2
=
z
1
+
z
2
=
z
1
+
z
2
⇒
|
1
z
1
+
1
z
2
|
=
|
z
1
+
z
2
|
(
∵
|
z
|
=
|
z
|
)
© examsnet.com
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