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Math Notes

Complex Numbers in Polar or Trigonometric Form, rcos(t) +risin(t), and Exponential Form, reit


In addition to representing complex numbers in Cartesian form as a + bi, we can also represent complex numbers in polar or trigonometric form.  The form gets its name because the form depends on polar coordinates containing an angle, the distance in radians from the positive x-axis, and a radius, the distance from the origin to the point in the plane.  It is sometimes called trigonometric form because the form of the complex number contains a cosine and a sine.  These forms of complex numbers are equivalent, but they serve different purposes.  We will come to what these are later on.

Image:Point in Polar coordinates.PNG

If we begin with a complex number in Cartesian form, we can determine it's polar form by calculating r and t, where r is the radius and t is the angle.

To find r for a complex number we, need to find the distance from the origin.  We can use the distance formula, and we get the value of  , so for the complex number z = 3 + 4i, .  We also sometimes see r indicated by the notation |z|, the modulus of z, or the length of z.  These notations are equivalent.

Then we need to find the angle t.  To do this, we can think of the complex number forming a right triangle with the x-axis.  From trigonometry, we can use the inverse tangent function to find the angle, given by   .  For the complex number 3 + 4i, the angle we get is tan-1(4/3) ≈ .93 radians or 53.1 degrees.  We should double check our answer to make sure our angle is in the right quadrant because it may be correct, or we may need to add π radians or 180 degrees to get it into the appropriate quadrant.  (Not sure which quadrant the complex number is in?  Graph it in the Cartesian plane.)  This angle is sometimes referred to as the argument of z or arg(z).

To get the polar form, we now use the same transformation that we would for any parametric representation of the x-y plane.  We set x = rcos(t) and y = rsin(t).  But remember, the x-coordinate here is the real part, and the y-coordinate here is the imaginary part, so a + bi becomes rcos(t) + risin(t).  Check our answer. 5cos(.93) = 2.99 and 5sin(.93) = 4.01.  This difference here is just a rounding error.

These forms of complex numbers also arise from another source, that is in exponential form.  Exponential form and polar form are closely related through the formula r*eit = r[cos(t) + isin(t)].  (This equation is referred to as the Euler equation.) This exponential form of complex numbers arise frequently in differential equations.  The properties of trigonometric forms of complex numbers can be proved using the exponential form as a starting point and applying well-known properties of exponents. It is through this Euler relationship that we know that the r and the t in both forms are the same, so the same calculations can be used to determine either form.  The exponential form of 3 + 4i, using our earlier results is 3 + 4i = 5e.93i.

The special properties of the polar form are derived from the properties of the exponential form.  For instance, what happens when we multiply two complex numbers in exponential form together? Like (2e.4i)(3e.6i).  We are going to multiply the coefficients and add the exponents (since the base is the same in both terms), giving us 6ei.  That was a lot easier than working with these numbers in Cartesian form.  Compare: 2e.4i = 2cos(.4) + 2isin(.4) = 1.84 + .779i, and 3e.6i = 3cos(.6) + 3isin(.6) = 2.48 + 1.69i; now multiply (1.84 + .779i)*(2.48 + 1.69i) by FOILing.  I think you see my point, but let's continue and compare answers.  4.56 + 3.11i + 1.93i + 1.32i2 = 3.24 + 5.04i.  What's 6ei? 6cos(1) + 6isin(1) = 3.24 + 5.05i.  The difference here is just a rounding error we could easily get rid of by carrying an extra digit.

Since exponential and polar form are the same, we can also represent this calculation in the following way. z1*z2 = r1*r2[cos(t1+t2) + isin(t1+t2)].  Just as with the exponential form, we are adding the angles (the exponent portion of the exponential form) and multiplying by the length or coefficient of the exponential form.  Just as we saw in the example above, the result is another calculation in polar or trigonometric form.  We can carry on this generalization for any number of complex numbers we wish to multiply.

Division is similarly much easier than in Cartesian form.  Consider .  When we do division of exponentials,

where before we added exponents, here we subtract them, and divide the coefficients, giving us 3e(π/2)i.  If we wanted to convert this now to Cartesian coordinates, we'd get 3cos(π/2) + 3isin(π/2) = 0 + 3i.  Compare that with

 There's no need to find the complex conjugate to solve this problem.  And what did we do?  Again, we can represent

 this process in polar form as z1/z2 = (r1/r2)[cos(t1 - t2) + isin(t1 - t2)].  Just as in the exponential form, we subtracted

 the angles, and divided the coefficients. [Note: 6cos(π) + 6isin(π) = 6(-1) + i(0) = -6; and 2cos(π/2) + 2isin(π/2) = 0

 + 2i(1) = 2i.]

One of the most powerful properties of the exponential and polar forms has to do with raising complex numbers to powers or taking roots of complex numbers.  Consider a simple exponential, like et.  If we wish to raise this to a large power, like (et)6, we multiply the exponent by the new power, giving us e6t.  This is the same process we use to determine large powers of complex numbers.  Consider the complex number 1 + i.  If we wanted to find (1 + i)6, we have two ways to approach this.  We can multiply it out in some way (either by hand or by applying the binomial theorem), or we can convert it into an exponential form (or polar form as we'll see later) and raise that to the sixth power.  Let's try it both ways.

Using the binomial theorem:

OR  and , thus (1+i)6 = .  Thus, , and

.  If we convert this back through polar form, we get 8cos(3π/2) + 8isin(3π/2) = 0 + 8i(-1) = -8i.

We can do this directly in polar form through the following equation zn = rn[cos(nt) + isin(nt)].  In words, raise the length of the complex number to the desired power, and multiply the angle by the desired power.

We can do roots very similarly, but where we multiply in exponential form for multiplication, we will now divide instead (since roots can be represented as fractional exponents, square root is a 1/2 exponent, cube root is a 1/3 exponent and so forth).  The rooting process is most useful for determining the roots of unity, or the roots of 1.

Consider the equation x5 = 1.  What are the values of x that make this work?  Well, we know one of them straight off the bat, 15 = 1, but this equation is equivalent to the expression x5 - 1 = 0.  We said before that because this is a fifth degree equation, there HAS to be five answers, five total values that, when we multiply the number by itself five times, we get a value of 1 at the end.  So far, we've found only one of these values, what are the other four?  To introduce our process, let's first start out by checking out our first value using exponential/polar form.  How can we write 1 + 0i as a complex exponential?  What is our length?  Well, clearly, it's 1.  What about our angle, well, it turns out to be 0, since our complex number is on the positive x-axis, and tan-1(0) = 0.  So, 1 = 1e0i.  To take the root of this number, the fifth root in this case, we are going to raise this number to the 1/5 power.  (1e0i)1/5 = 11/5*e0i/5 = 1e0i = 1.  We get back to where we started, at 1, because 1 to any power is 1, and 0/5 is still 0.

The thing about angles, though, is that eventually you come around full circle if your angle is big enough.  Big enough, in radians, is 2π.  It's still true if you add another 2π to get 4π, and another 2π to get 6π and so on.  In angular terms 0 = 2π = 4π = 6π = 8π.  But it's exactly these angles that are going to get us our other four roots.  Redo the calculation above, but replace 0i with 2πi.  (1e2πi)1/5 = 11/5*ei/5 = 1e2πi/5 = cos(2π/5) + isin(2π/5) = .3090 + .9511i.  (If you can run your calculator in complex number mode, you can check it out, raise this number to the fifth power and you get 1, especially if you carry several digits.)  We can repeat this process for the other angles.  (1e4πi)1/5 = 11/5*ei/5 = 1e4πi/5 = cos(4π/5) + isin(4π/5) = -.8090 + .5878i(1e6πi)1/5 = 11/5*ei/5 = 1e6πi/5 = cos(6π/5) + isin(6π/5) = -.8090 - .5878i.   (1e8πi)1/5 = 11/5*ei/5 = 1e8πi/5 = cos(8π/5) + isin(8π/5) = .3090 - .9511i.  Those are my five roots, and notice that they come in complex conjugate pairs, just as we would have expected.  We could continue like this, but if we do, we will just start to repeat the numbers we already have.

To describe this procedure directly in polar form we might write  , where

again we are taking just the primary root of the length of the complex number, and we are dividing the angle by number of the root.  The 2kπ term refers to the process of adding multiples of 2π and repeating the process until it either starts to repeat, or we achieve the desired number of roots.

It should be noted that polar and exponential forms are less useful for adding and subtracting complex numbers.  Convert to Cartesian form to perform these operations, and then convert back if need be.

Worked Examples

A. Find the polar form and exponential form of the following complex numbers:

a. 1 + i                         b. -i                     c.  3                       d. -2 - 3i

For a, we find |z| = , and  , thus 1 + i = cos(π/4) + isin(π/4) = .

For b, we find |z| = 1 and t = tan-1(1/0).  Tangent is undefined for the angle π/2, thus this is either that or that plus π, and since -i is along the negative y-axis, this is the second choice, 3π/2.  Thus polar form is cos(3π/2) + isin(
3π/2) and exponential form is e3π/2i.
For c, the length is clearly 3, and the angle is 0 since 3 is on the positive x-axis.  The polar form is 3cos(0) + 3isin(0), and the exponential form is 3e0i.

For d, we find |z| = .  And the angle is , but that angle is not in the third quadrant where our

complex number is, so we add 3.14159 to get 4.12.  Thus, the polar form is cos(4.12) +
isin(4.12), and the

exponential form is

B. Multiply. a.  (2e.54i)(3.4e-.89i)                 b. 4[cos(.12) + isin(.12)]*3[cos(-.7) + isin(-.7)]

Multiply the coefficients and add the angles/exponents.  For a, 2*3.4 = 6.8 and .54 + -.89 = -.35; thus (2e.54i)(3.4e-.89i) = 6.8e-.35i.  For b, 4*3 = 12 and .12 + -.7 = -.58; thus 4[cos(.12) + isin(.12)]*3[cos(-.7) + isin(-.7)] = 12[cos(-.58) + isin(-.58)].


C. Divide. a.            b.

Divide the coefficients, and subtract the angles/exponents.  For a, 4/2 = 2, 3π/2 - π/4 = 5π/4; thus .  For b, 6/3 = 2 and π/6 - π/2 = -π/3; thus  .


D. Simplify.  (2 - i)5.  State your answer in Cartesian form.

First convert 2 - i either to polar or exponential form.  |z| = .  And .  Check the quadrant.  2 - i, is the point (2, -1) in the complex plane, that's in the fourth quadrant, and so is -1.11 radians, so we don't have to add any angles.  The exponential form for 2 - i is and the polar form is .  In order raise these numbers to the fifth power, we raise the coefficient to the fifth power, and then multiply the angle by 5.  , and -.464 = -2.32.  We should reduce our angles as much as possible.  Generally, it's okay to use negative angles as long as they are less than π, though preferably, π/2.  When they get bigger in magnitude than that, we want to, in this case, add 2π.  -2.32 + 2π = 3.96.  We would do the same thing if we had a positive angle larger than 2π, but then we would subtract to reduce the angle, as many times as it necessary.  Thus (2 - i)5 = and in polar form. We can now find out what this is in Cartesian form by evaluating the polar form:  38.07 + 40.93i.  If you have a calculator than can check the calculation for you, you get 38 + 41i.

E. Simplify.  State your answer in Cartesian form.

First we need to covert 2 - i to exponential or polar form.  We did this in the previous worked example.  2 - i = = .  To find the fourth root, we need to take the fourth root of the coefficient, and divide the angle/exponent by four.  and -.464/4 = -.116.  Thus .  To get the Cartesian form, evaluate the polar form: 1.21 - .142i.  If your calculator has the ability to work with complex numbers, you can check this by entering (2 - i)^(1/4) = 1.21 - .141i

Problem Solving Tips

  • For problems involving multiplication and division, use whichever method you feel most comfortable with.  If you are given Cartesian form, and like that method, then use that method.  If you are familiar with the binomial theorem, you can use that method to get around converting to exponential form as well--although, remember you have to simplify and reduce all powers of i.
  • You have to find two values regardless of whether or not you use polar or exponential form.  Those values are the modulus of the complex number, and the angle.
  • The modulus depends on the Cartesian form.  It's the distance from the origin to the point in the complex plane that represents the complex number in question.  You can also think of it as the hypotenuse of a right triangle with side a and b.
  • You have to use the exponential or polar form to solve for roots of unity or roots in general.  You can't use the Cartesian form for this.
  • For roots of unity, remember that you are looking for the same number of roots as the power of the equation.  There are five fifth roots, and six sixth roots, and so on.  To find each root, add 2π and repeat the calculation.
  • When calculating polar or exponential form, make sure that you check to see that your angle is in the correct quadrant.  You may need to add π.
  • When using the polar or exponential form for multiplication, you may need to add/subtract multiples of 2π to keep the angle generally between -π and 2π.
  • When doing roots of unity, check that your complex answers come in pairs of complex conjugates.

Additional Problems with Answers

1. Find the polar form of the following complex numbers; state the modulus and the argument:

    a. 1 - i             b. -2               c. -3 + i            d. 4 + 6i

2. Multiply the following complex numbers as directed, report your answers in polar form:

    a. (1 + i)(2 - i)           b. (3e-4i)(6e3i)            c. 2[cos(.4) + isin(.4)]*0.1[cos(2.3) + isin(2.3)]      d. (3 - i)(2e.01i)

3. Divide the following complex numbers as directed, report your answers in exponential form:

    a.             b.            c.           d.

4. Simplify.  State your answer in Cartesian form.

    a.  (3 + 3i)7           b. (1 - 2i)4        c. (5 + 3i)8

5. Simplify.

    a.              b.          c.

6. Find the following roots of unity:


1. a. |z| =, arg(z) = -π/4, [cos(π/4) +isin(π/4)]
    b. |z| = 2, arg(z) = π, 2[cos(π)+isin(π)]
    c. |z| =
, arg(z) = -.32175, [cos(-.32175) + isin(-.32175)]
    d. |z| = , arg(z) = .98279, [cos(.98279) + isin(.98279)]
a. [cos(-.32175) + isin(-.32175)]
    b. 18[cos(-1) + isin(-1)]
    c. 0.2[cos(2.7) + isin(2.7)]
    d. [cos(-.31175) + isin(-.31175)]
3. a.




4. a. 17496 - 17496i
    b. -7 + 24i
    c. -506864 - 1236480i
5. a. 1.598 - .13080i
    b. 2 + i
    c. 1.5776 - .28730i
6. a. 1, -1, i, -i
    b. 1, 1/2 + .866i, -1/2 + .866i, -1, -1/2 - .866i, 1/2 - .866i
    c. 1, .62349 + .78183i, -.2225 + .97497i, -.90097 + .43388i, -.90097 - .43388i, -.2225 - .97497, .62349 - .78183i
   d. 1, .766 + .64279i, .17365 + .9848i, -1/2 + .866i, -.9397 + .342i, -.9397 - .342i, -1/2 - .866i, .17365 - .9848i, .766 - .64279i.

Links to Outside Sources

Lesson on Complex Numbers or Imaginary Numbers
Complex Number -- from MathWorld
About.com Complex Numbers
The polar form for complex numbers
Summary of complex number operations
Exponential Function and Complex Numbers
Module 4: Polar and Exponential Form
Complex Numbers

Links to Supporting Topics

Imaginary Numbers
Complex Numbers in Cartesian Form (a + bi)
Math Forum: Ask Dr. Math FAQ: Imaginary Numbers
Answers and Explanations -- Do "Imaginary Numbers" Really Exist?
Complex numbers - Wikibooks
College Algebra Tutorial on Complex Numbers
Complex numbers - A complete course in algebra
Complex Numbers and Trigonometry


Question Corner -- The Origin of Complex Numbers
A History of Hypercomplex Numbers
Re: so, who invented the IMAGINARY NUMBER?
Wikipedia article


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