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MA 1165 - Lecture 25
4/17/09
1
Polar Form
We’ve expressed complex numbers in a basic form, like
a + bi,
(1)
r(cos(θ) + i sin(θ)),
(2)
and in a form using cosines and sines, like
which I will now call polar form.
It’s relatively easy to switch from polar form to the basic form. We just evaluate the cosine and sine at θ,
and simplify. For example, if r = 2 and θ = π4 , we get
√
√
√
√
2(cos( π4 ) + i sin( π4 )) = 2( 22 + 22 i ) = 2 + 2 i.
(3)
Even if the angle is not a nice one, we can just approximate. Let’s do this one in degrees.
3(cos(40◦) + i sin(40◦ )) = 3(0.7660 + 0.6428 i) = 2.2981 + 1.9284 i.
(4)
imaginary axis
r(cos(θ) + i sin(θ))
b
r
θ
real axis
a
Figure 1: Converting a + bi into polar form.
Converting in the other direction is too bad either. It’s maybe not as obvious, however. Suppose we have a
complex number a + bi, and we want to find r and θ (see Figure 1). We can find r using the Pythagorean
theorem,
r 2 = a2 + b 2 ,
(5)
or directly with
r=
p
a2 + b 2 .
(6)
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MA 1165 - Lecture 25
Once we have r, we know that a = r cos(θ) and b = r sin(θ), so we can find θ with these. It’s a bit easier to
find these another way, however. If you look at Figure 1, the side opposite θ is b, and the side adjacent is a.
The tangent function is defined in terms of these two quantities. We have, therefore,
tan(θ) =
b
.
a
(7)
We can solve this equation, but we’ll always end up with
b
−1
.
θ = tan
a
(8)
As an example, suppose we have the complex number
2 + 3i.
(9)
To put this in polar form, we just plug into formulas (6) and (8). In particular, we get
p
p
√
√
r = a2 + b2 = 22 + 32 = 4 + 9 = 13,
and
θ = tan−1
b
a
= tan−1
3
2
(10)
= 56.32◦
(11)
√
13(cos(56.32◦) + i sin(56.32◦)).
(12)
in degrees. In polar form, we have
2 + 3i =
2
Quiz 25A
Convert the following complex numbers into polar form. Choose your answers from
√
√
(a) 2( cos(45◦ ) + i sin(45◦))
(b) 1( cos(180◦ ) + i sin(180◦))
(c) 5( cos(26.57◦) + i sin(26.57◦))
(d) 1( cos(90◦) + i sin(90◦ ))
(e) none of these
QA1.
1 + i.
QA2.
2 + i.
QA3.
i.
QA4.
−1.
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DeMoivre’s Theorem
Why would we want to put a complex number into polar form? There are lots of reasons. Here’s one.
We squared complex numbers last time. That’s not too bad, but you can’t just put it into your calculator
(actually, some of you can). Think about computing (2 − 5i)5 . This would be somewhat complicated. In
polar form, this turns out to be fairly simple. To do this, we need a somewhat surprising theorem.
DeMoivre’s Theorem 1. For a complex number in polar form, we have
( r(cos(θ) + i sin(θ) )n = r n (cos(nθ) + i sin(nθ)).
(13)
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MA 1165 - Lecture 25
For example, suppose we had the following computation to perform.
5
( 2(cos(20◦ ) + i sin(20◦ )) ) .
(14)
DeMoivre’s theorem tells us that this is equal to
5
( 2(cos(20◦ ) + i sin(20◦)) ) = 25 (cos(5 · 20◦) + i sin(5 · 20◦ )) = 32(cos(100◦) + i sin(100◦ )).
4
(15)
Quiz 25B
Apply DeMoivre’s theorem, and then convert to basic form. Choose your answers from
(a) 9
(b) 64i
(c) 64
(d) i
6
QB1.
( 2( cos(60◦) + i sin(60◦ )) ) .
QB2.
( 1( cos(90◦) + i sin(90◦ )) ) .
QB3.
( 3( cos(180◦) + i sin(180◦ )) ) .
5
(e) none of these
5
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n-th roots
Computing square roots, cube roots, and n-th roots in general, is just as easy as it is for real numbers. We
just use DeMoivre’s theorem in reverse. Let’s start with something easy. The cube roots of 1. For complex
numbers, there are always three cube roots, four fourth roots, etc. Therefore, in the complex numbers, 1
should have three cube roots.
In polar form,
1 = 1( cos(0◦ ) + i sin(0◦ )).
(16)
We take the cube root of the 1 outside, and then divide the angle by three
√
√
3
3
1 = 1( cos(0◦ /3) + i sin(0◦ /3)) = 1( cos(0◦ ) + i sin(0◦ )).
(17)
The angle for 1 can also be 360◦ and 720◦ and any other multiple of 360◦ , so we can get cube roots of 1 with
√
3
1( cos(360◦ /3) + i sin(360◦ /3)) = 1( cos(120◦ ) + i sin(120◦)),
(18)
and
√
3
1( cos(720◦ /3) + i sin(720◦ /3)) = 1( cos(240◦ ) + i sin(240◦)).
(19)
The other multiples of 360◦ will give these same three complex numbers. If we insert the values of the cosines
and sines, our three cube roots of 1 are
1,
− 12 +
√
3i
2 ,
and − 12 −
√
3i
2
(20)
The cube roots of 5 are computed
the same way. Since the angles will be the same, the only difference will
√
be that everything will be 3 5 larger. The cube roots of 3, therefore, will be
√
√
√
√
√
3
3 √
3
3 √
3
(21)
5,
− 25 + 52 3 i , and − 25 − 52 3 i
Without DeMoivre’s theorem, it would be difficult to imagine what the cube roots of i would be. But this
is pretty easy. Just write i in polar form three different ways, and do what we did above.
i = 1( cos(90◦ ) + i sin(90◦ )) = 1( cos(450◦) + i sin(450◦ )) = 1( cos(810◦) + i sin(810◦ ))
(22)
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MA 1165 - Lecture 25
are three ways of writing i in polar form. Take the cube root of 1 and divide the angles by three to get
1( cos(30◦ ) + i sin(30◦)),
1( cos(150◦) + i sin(150◦ )), and 1( cos(270◦) + i sin(270◦ )).
(23)
Computing the cosines and sines gives us the three cube roots of i in basic form.
√
3
2
6
+ 2i ,
−
√
3
2
+ 2i , and
(24)
0 − i = −i.
Quiz 25C
For each of the following, one of the roots is in the list of answers, choose that as your answer.
(a)
1
2
−
√
3i
2
(b)
√
3
2
+
QC1.
The cube roots of −1.
QC2.
The fourth roots of 1.
QC3.
The square roots of 1.
7
i
2
(c) −
√
3
2
−
i
2
(d) 1
(e) none of these
Homework 25
For problems 1-4, convert the following complex numbers into polar form. Choose your answers from
◦
◦
(a) 5(
))
(b) 3( cos(48.19◦) + i sin(48.19◦))
√ cos(36.87◦ ) + i sin(36.87
◦
(d) 8( cos(45 ) + i sin(45 ))
(e) none of these
1.
2.
(c) 1( cos(270◦ ) + i sin(270◦))
2 + 2i.
√
2 + 5 i.
3.
−i.
4.
4 + 3i.
For problems 5-7, apply DeMoivre’s theorem, and then convert to basic form. Choose your answers from
(a) −0.6840 − 1.8794i
(b) 27
(c) −10.94 − 30.07i
(d) 0.1736 + 0.9848i
(e) none of these
8
5.
( 1( cos(10◦) + i sin(10◦ )) ) .
6.
( 3( cos(120◦) + i sin(120◦ )) ) .
7.
( 2( cos(50◦) + i sin(50◦ )) )5 .
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For problems 8-10, one of the roots is in the list of answers, choose that as your answer.
(a) 1
(b)
√
3
2
−
i
2
(c) −
8.
The cube roots of −i.
9.
The fourth roots of −1.
10.
The square roots of
1
2
+
√
2
2
√
3i
2 .
−
√
2i
2
(d)
√
3
2
+
i
2
(e) none of these