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Byju's Answer
Standard XII
Mathematics
Special Integrals - 1
Convert the f...
Question
Convert the following in the polar form:
(i)
1
+
7
i
(
2
−
i
)
2
(ii)
1
+
3
i
1
−
2
i
Open in App
Solution
(i) Here,
z
=
1
+
7
i
(
2
−
i
)
2
=
1
+
7
i
4
+
i
2
−
4
i
=
1
+
7
i
4
−
1
−
4
i
=
1
+
7
i
3
−
4
i
×
3
+
4
i
3
+
4
i
=
3
+
4
i
+
21
i
+
28
i
2
3
2
+
4
2
=
3
+
4
i
+
21
i
−
28
3
2
+
4
2
=
−
25
+
25
i
25
=
−
1
+
i
Let
r
cos
θ
=
−
1
and
r
sin
θ
=
1
On squaring and adding, we obtain
r
2
(
cos
2
θ
+
sin
2
θ
)
=
1
+
1
⇒
r
2
(
cos
2
θ
+
sin
2
θ
)
=
2
⇒
r
2
=
2
(
∵
cos
2
θ
+
sin
2
θ
=
1
)
⇒
r
=
√
2
(As r>0 )
∴
√
2
cos
θ
=
−
1
and
√
2
sin
θ
=
1
⇒
cos
θ
=
−
1
√
2
and
sin
θ
=
1
√
2
∴
θ
=
π
−
π
4
=
3
π
4
(As
θ
lies in II quadrant )
∴
z
=
r
cos
θ
+
i
r
sin
θ
=
√
2
cos
3
π
4
+
i
√
2
sin
3
π
4
=
√
2
(
cos
3
π
4
+
i
sin
3
π
4
)
This is the required polar form.
(ii) Here,
z
=
1
+
3
i
1
−
2
i
=
1
+
3
i
1
−
2
i
×
1
+
2
i
1
+
2
i
=
1
+
2
i
+
3
i
−
6
1
+
4
=
−
5
+
5
i
5
=
−
1
+
i
Let
r
cos
θ
=
−
1
and
r
sin
θ
=
1
On squaring and adding, we obtain
r
2
(
cos
2
θ
+
sin
2
θ
)
=
1
+
1
⇒
r
2
(
cos
2
θ
+
sin
2
θ
)
=
2
⇒
r
2
=
2
(
∵
cos
2
θ
+
sin
2
θ
=
1
)
⇒
r
=
√
2
(As r>0 )
∴
√
2
cos
θ
=
−
1
and
√
2
sin
θ
=
1
⇒
cos
θ
=
−
1
√
2
and
⇒
sin
θ
=
1
√
2
∴
θ
=
π
−
π
4
=
3
π
4
(As
θ
lies in II quadrant )
∴
z
=
r
cos
θ
+
i
r
sin
θ
=
√
2
cos
3
π
4
+
i
√
2
sin
3
π
4
=
√
2
(
cos
3
π
4
+
i
sin
3
π
4
)
This is the required polar form.
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