14.3 The trigonometric form of a complex number

When we say z:=x+iyz:=x+iy (also known as \sayz is defined as x+iyx+iy), xx and yy can be anything! For example, xx and yy could equal cos(θ)\cos(\theta) and sin(θ)\sin(\theta) respectively. By setting different values of theta, we can obtain any point on a circle with radius 1, and centre (0,0)(0,0). For example, if we have z:=cos(θ)+isin(θ)z:=\cos(\theta)+i\sin(\theta), then to obtain the number 11, we can just set θ=0\theta=0. To be able to obtain every complex number, however, we need to introduce a second variable (which we can call rr). This \sayscales the circle, so that for each value of rr, the values of θ\theta between 0 and 2π2\pi (2π2\pi is not inclusive) we can obtain a circle with that radius. Overall, we can write any complex number in the form

z:=r(cos(θ)+isin(θ)).z:=r\left(\cos(\theta)+i\sin(\theta)\right). (14.36)

Additionally, we can do this with r0r\geqq 0 and πθπ-\pi\leqq\theta\leqq\pi.

To find the trigonometric form of a given complex number there are two ways.

To convert a complex number (e.g. 1+3i1+3i) into trigonometric form, the first way algebra is to use algebra (in particular \saycomparing coefficients). By comparing rcos(θ)+risin(θ)r\cos(\theta)+ri\sin(\theta) with 1+3i1+3i, we obtain that rcos(θ)=1r\cos(\theta)=1 and that rsin(θ)=3r\sin(\theta)=3. Thus

r2cos2(θ)+r2sin2(θ)\displaystyle r^{2}\cos^{2}(\theta)+r^{2}\sin^{2}(\theta) =r2(cos2(θ)+sin2(θ)\displaystyle=r^{2}(\cos^{2}(\theta)+\sin^{2}(\theta)
=2 (from 1+3i)\displaystyle=2\text{ (from $1+3i$)}

from which we can deduce that r=±2r=\pm\sqrt{2}. The other thing we can do is divide through, thus obtaining that rsin(θ)rcos(θ)=11\frac{r\sin(\theta)}{r\cos(\theta)}=\frac{1}{1}, and thus that θ=arctan(1)\theta=\arctan(1) (which is 22\frac{\sqrt{2}}{2}). Overall, we can then write that

1+i=2e22i1+i=\sqrt{2}e^{\frac{\sqrt{2}}{2}i} (14.37)

which we can also do for any complex number.

The second way involves geometry (I still need to find where I originally wrote my notes on this, but the method is to draw the complex number - not necessary, but usually helpful - and to then find the modulus and argument of the complex number). First we can apply this useful fact:

Theorem 14.3.1

For a complex number z=r(cos(θ)+isin(θ))z=r(\cos(\theta)+i\sin(\theta)), arg(z)=θ\arg(z)=\theta and |z|=r|z|=r

The proof is definitely not relevant for any A Level examination, but it’s also not too difficult.

  • First we prove that arg(z)=θ\arg(z)=\theta. We know that for a complex number x+iyx+iy, there are several cases for the argument (rather unfortunately we need to consider each quadrant separately)

    • In the first case, x,y0x,y\geqq 0, where arg(z)=arctan(yx)\arg(z)=\arctan\left(\frac{y}{x}\right). For our zz, this means

      arg(z)\displaystyle\arg(z) =arctan(rsin(θ)rcos(θ))\displaystyle=\arctan\left(\frac{r\sin(\theta)}{r\cos(\theta)}\right) (14.38)
      =arctan(tan(θ))\displaystyle=\arctan(\tan(\theta)) (14.39)
      =θ\displaystyle=\theta (14.40)
    • TODO: go through the other three cases

  • Second we prove that |z|=r|z|=r. This proof is quite a bit more satisfying than the previous proof, we just apply the definition of the modulus, that is that

    z2\displaystyle z^{2} =x2+y2\displaystyle=\sqrt{x^{2}+y^{2}} (14.41)
    =((rcos(θ))2+(rsin(θ))2)\displaystyle=\sqrt{\Big{(}\big{(}r\cos(\theta)\big{)}^{2}+\big{(}r\sin(\theta% )\big{)}^{2}\Big{)}} (14.42)
    =r2(cos2(θ)+sin2(θ))\displaystyle=\sqrt{r^{2}\big{(}\cos^{2}(\theta)+\sin^{2}(\theta)\big{)}} (14.43)
    =r2\displaystyle=\sqrt{r^{2}} (14.44)

    Note that we also defined r0r\geqq 0 (well, hopefully we did) and thus

    r2=r\sqrt{r^{2}}=r (14.45)

14.3.1 Using the trigonometric form of a complex number

Example 14.3.1

This is probably not the most exciting example, but we might want to consider the case of

z=2(cos(330)sin(330))z=2(\cos(330^{\circ})-\sin(330^{\circ})) (14.46)

If we can write this in trigonometric form, then by 14.3.1 we can just read the argument and modulus. Examining the expression, it looks pretty close to something in the form

z=r(cos(θ)+sin(θ)).z=r(\cos(\theta)+\sin(\theta)). (14.47)

The only problem we have is that there’s a pesky negative hanging around in there, but not to worry, we can apply two useful facts here; cos(θ)=cos(θ)\cos(\theta)=\cos(-\theta) and sin(θ)=sin(θ)\sin(\theta)=-\sin(-\theta) (these two equations are referred to as the \sayeven and \sayodd property of cos\cos and sin\sin respectively - see Section 14.7.3 for a proof of this).

Therefore, we can rewrite zz

z\displaystyle z =2(cos(330)sin(330))\displaystyle=2(\cos(330^{\circ})-\sin(330^{\circ})) (14.48)
=2(cos(330)+sin(330))\displaystyle=2(\cos(-330^{\circ})+\sin(-330^{\circ})) (14.49)
=2(cos(30)+sin(30))\displaystyle=2(\cos(-30^{\circ})+\sin(-30^{\circ})) (14.50)

Therefore the argument is 30=π6 rad-30^{\circ}=-\frac{\pi}{6}\text{ rad}