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Byju's Answer
Standard XI
Mathematics
Differentiability in an Interval
The amplitude...
Question
The amplitude of
1
i
is equal to
(a) 0
(b)
π
2
(c)
-
π
2
(d) π
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Solution
(c)
-
π
2
Let
z
=
1
i
⇒
z
=
1
i
×
i
i
⇒
z
=
i
i
2
⇒
z
=
-
i
Since
,
z
0
,
-
1
lies
on
the
negative
imaginary
axis
.
Therefore
,
arg
(
z
)
=
-
π
2
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