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Question #349054
Convert 1−i/2 into trigonometric form
Expert's answer
r
2
=
x
2
+
y
2
=
(
1
)
2
+
(
−
1
2
)
2
=
5
4
r^2=x^2+y^2=(1)^2+(-\dfrac{1}{2})^2=\dfrac{5}{4}
r
2
=
x
2
+
y
2
=
(
1
)
2
+
(
−
2
1
)
2
=
4
5
tan
θ
=
−
1
/
2
1
=
−
1
2
\tan \theta=\dfrac{-1/2}{1}=-\dfrac{1}{2}
tan
θ
=
1
−
1/2
=
−
2
1
Quadrant IV
r
=
5
2
,
θ
=
2
π
−
tan
−
1
1
2
r=\dfrac{\sqrt{5}}{2}, \theta=2\pi-\tan^{-1}\dfrac{1}{2}
r
=
2
5
,
θ
=
2
π
−
tan
−
1
2
1
z
=
5
2
(
cos
(
2
π
−
tan
−
1
1
2
)
+
i
sin
(
2
π
−
tan
−
1
1
2
)
)
z=\dfrac{\sqrt{5}}{2}(\cos(2\pi-\tan^{-1}\dfrac{1}{2})+i\sin(2\pi-\tan^{-1}\dfrac{1}{2}) )
z
=
2
5
(
cos
(
2
π
−
tan
−
1
2
1
)
+
i
sin
(
2
π
−
tan
−
1
2
1
))
Or
r
=
−
5
2
,
θ
=
π
−
tan
−
1
1
2
r=-\dfrac{\sqrt{5}}{2}, \theta=\pi-\tan^{-1}\dfrac{1}{2}
r
=
−
2
5
,
θ
=
π
−
tan
−
1
2
1
z
=
−
5
2
(
cos
(
π
−
tan
−
1
1
2
)
+
i
sin
(
π
−
tan
−
1
1
2
)
)
z=-\dfrac{\sqrt{5}}{2}(\cos(\pi-\tan^{-1}\dfrac{1}{2})+i\sin(\pi-\tan^{-1}\dfrac{1}{2}) )
z
=
−
2
5
(
cos
(
π
−
tan
−
1
2
1
)
+
i
sin
(
π
−
tan
−
1
2
1
))
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