z=√3-i, find its polar form
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Step-by-step explanation:
z=√3-i, find its polar form is Explanation:
Only for real
a
2
, sqrta^2)=|a| #, by convention ( or definition ).
So, the other square root
−
|
a
|
is out of reach.
Here, for sqrt(z), where z is complex, we cannot conveniently take
one and keep off the other.
So,
√
3
+
4
i
=
(
3
+
i
4
)
1
2
that has two values.
They are
√
5
(
cos
a
+
i
sin
a
)
1
2
,
where
cos
a
=
3
5
and
sin
a
=
4
5
=
√
5
(
cos
(
a
+
2
k
π
)
+
i
sin
(
a
+
2
k
π
)
)
1
2
,
k
=
0
,
1
√
5
(
cos
(
a
2
+
k
π
)
+
i
sin
(
a
2
+
k
π
)
)
,
k
=
0
,
1
,
using De Moivre's theorem
=
√
5
(
cos
(
a
2
)
+
i
sin
(
a
2
)
)
and
√
5
(
cos
(
a
2
+
π
)
+
i
sin
(
a
2
+
π
)
)
=
±
√
5
(
cos
(
a
2
)
+
i
sin
(
a
2
)
)
=
√
5
c
i
s
(
26.565
o
)
, nearly.
using
cos
(
π
+
θ
)
=
−
cos
θ
and
sin
(
π
+
θ
)
=
−
sin
θ
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