3.3 Powers and roots

Example 1

If \displaystyle z = \frac{1+i}{\sqrt2}, determine \displaystyle z^3 and \displaystyle z^{100}.

We write \displaystyle z in polar form \displaystyle \ \ z= \frac{1}{\sqrt2} + \frac{i}{\sqrt2} = 1\cdot \Bigl(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\Bigr)\ \ and Moivre's formula gives

Example 2

In the usual way one does an expansion by means of the squaring rules

and according to de Moivre's formula one gets

If one equates the real and imaginary parts of the two expressions one gets the well-known trigonometric formulas

Example 3

Simplify \displaystyle \ \ \frac{(\sqrt3 + i)^{14}}{(1+i\sqrt3\,)^7(1+i)^{10}}\,.

We write the numbers \displaystyle \sqrt{3}+i, \displaystyle 1+i\sqrt{3} and \displaystyle 1+i in polar form

• \displaystyle \quad\sqrt{3} + i = 2\Bigl(\cos\frac{\pi}{6} + i\,\sin\frac{\pi}{6}\,\Bigr)\vphantom{\biggl(},
• \displaystyle \quad 1+i\sqrt{3} = 2\Bigl(\cos\frac{\pi}{3} + i\,\sin\frac{\pi}{3}\,\Bigr)\vphantom{\biggl(},
• \displaystyle \quad 1+i = \sqrt2\,\Bigl(\cos\frac{\pi}{4} + i\,\sin\frac{\pi}{4}\,\Bigr)\vphantom{\biggl(}.

Then we get with de Moivre's formula

and this expression can be simplified by performing multiplication and division in polar form

Exempel 4

Solve the binomial equation \displaystyle \ z^4= 16\,i\,.

Write \displaystyle z and \displaystyle 16\,i in polar form

• \displaystyle \quad z=r\,(\cos \alpha + i\,\sin \alpha)\,,
• \displaystyle \quad 16\,i= 16\Bigl(\cos\frac{\pi}{2} + i\,\sin\frac{\pi}{2}\,\Bigr)\vphantom{\biggl(}.

This turns the equation \displaystyle \ z^4=16\,i\ into

Matching the moduli and arguments on both sides gives

Example 5

For a real number \displaystyle z the definition is consistent with the case when the exponent is real, as \displaystyle z=a+0\cdot i which gives

Example 6

A further indication of why the above definition is so natural is given by the relationship

which demonstrates that de Moivre's formula is actually identical to the well-known law of exponents,

Example 7

From the above definition, one can obtain the relationship

which connects together the, generally regarded, most basic numbers in mathematics: \displaystyle e, \displaystyle \pi, \displaystyle i and 1. This relationship is seen by many as the most beautiful in mathematics and was discovered by Euler in the early 1700's.

Example 8

Solve the equation \displaystyle \ (z+i)^3 = -8i.

Put \displaystyle w = z + i. We then get the binomial equation \displaystyle \ w^3=-8i\,. To begin with, we rewrite \displaystyle w and \displaystyle -8i in polar form

• \displaystyle \quad w=r\,(\cos \alpha + i\,\sin \alpha) = r\,e^{i\alpha}\,\mbox{,}
• \displaystyle \quad -8i = 8\Bigl(\cos \frac{3\pi}{2} + i\,\sin\frac{3\pi}{2}\,\Bigr) = 8\,e^{3\pi i/2}\vphantom{\biggl(}\,\mbox{.}

The equation in polar form is \displaystyle \ r^3e^{3\alpha i}=8\,e^{3\pi i/2}\ and matching the moduli and arguments on both sides gives,

The roots of the equation are thus

• \displaystyle \quad w_1 = 2\,e^{\pi i/2} = 2\Bigl(\cos \frac{\pi}{2} + i\,\sin\frac{\pi}{2}\,\Bigr) = 2i\,\mbox{,}\quad\vphantom{\biggl(}
• \displaystyle \quad w_2 = 2\,e^{7\pi i/6} = 2\Bigl(\cos\frac{7\pi}{6} + i\,\sin\frac{7\pi}{6}\,\Bigr) = -\sqrt{3}-i\,\mbox{,}\quad\vphantom{\Biggl(}
• \displaystyle \quad w_3 = 2\,e^{11\pi i/6} = 2\Bigl(\cos\frac{11\pi}{6} + i\,\sin\frac{11\pi}{6}\,\Bigr) = \sqrt{3}-i\,\mbox{,}\quad\vphantom{\biggl(}

i.e. \displaystyle z_1 = 2i-i=i, \displaystyle z_2 = - \sqrt{3}-2i and \displaystyle z_3 = \sqrt{3}-2i.

Example 9

Solve \displaystyle \ z^2 = \overline{z}\,.

If for \displaystyle z=a+ib one has \displaystyle |\,z\,|=r and \displaystyle \arg z = \alpha then for \displaystyle \overline{z}= a-ib one gets \displaystyle |\,\overline{z}\,|=r and \displaystyle \arg \overline{z} = - \alpha.This means that \displaystyle z=r\,e^{i\alpha} and \displaystyle \overline{z} = r\,e^{-i\alpha}. The equation can be written

which directly gives that \displaystyle r=0 is a solution, i.e. \displaystyle z=0. If we assume that \displaystyle r\not=0 then the equation can be written as \displaystyle \ r\,e^{3i\alpha} = 1\,, which gives after matching moduli and arguments

The solutions are

• \displaystyle \quad z_1 = e^0 = 1\,\mbox{,}
• \displaystyle \quad z_2 = e^{2\pi i/ 3} = \cos\frac{2\pi}{3} + i\,\sin\frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt3}{2}\,i\,\mbox{,}\vphantom{\Biggl(}
• \displaystyle \quad z_3 = e^{4\pi i/ 3} = \cos\frac{4\pi}{3} + i\,\sin\frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt3}{2}\,i\,\mbox{,}
• \displaystyle \quad z_4 = 0\,\mbox{.}

Example 10

1. Solve the equation \displaystyle \ x^2-6x+7=2\,.
   
   The coefficient in front of \displaystyle x is \displaystyle -6 and it shows that we must have the number \displaystyle (-3)^2=9 as the constant term on the left-hand side to make a complete square. By adding \displaystyle 2 to both sides we achieve this:

   \displaystyle \begin{align*} x^2-6x+7+2 &= 2+2\,\mbox{,}\\ x^2-6x+9\phantom{{}+2} &= 4\,\mbox{,}\\ \rlap{(x-3)^2}\phantom{x^2-6x+7+2}{} &= 4\,\mbox{.}\end{align*}

   Taking roots then gives \displaystyle x-3=\pm 2, which means that \displaystyle x=1 or \displaystyle x=5.

2. Solve the equation \displaystyle \ z^2+21=4-8z\,.
   
   The equation can be written as \displaystyle z^2+8z+17=0. By subtracting 1 on both sides, we get a complete square on the left-hand side:

   \displaystyle \begin{align*} z^2+8z+17-1 &= 0-1\,\mbox{,}\\ z^2+8z+16\phantom{{}-1} &= -1\,\mbox{,}\\ \rlap{(z+4)^2}\phantom{z^2+8z+17-1}{} &= -1\,\mbox{,}\end{align*}

   and thus \displaystyle z+4=\pm\sqrt{-1}. In other words, the solutions are \displaystyle z=-4-i and \displaystyle z=-4+i.

Example 11

Solve the equation \displaystyle \ x^2-\frac{8}{3}x+1=2\,.

Half the coefficient of \displaystyle x is \displaystyle -\tfrac{4}{3}. We thus add \displaystyle \bigl(-\tfrac{4}{3}\bigr)^2=\tfrac{16}{9} to both sides

Now it's easy to get to \displaystyle x-\tfrac{4}{3}=\pm\tfrac{5}{3} and thus to get that \displaystyle x=\tfrac{4}{3}\pm\tfrac{5}{3}, i.e. \displaystyle x=-\tfrac{1}{3} or \displaystyle x=3.

Example 12

Solve the equation \displaystyle \ x^2+px+q=0\,.

Completing the square gives

This gives the usual formula, pq-formula, for solutions to quadratic equations

Example 13

Solve the equation \displaystyle \ z^2-(12+4i)z-4+24i=0\,.

Half the coefficient of \displaystyle z is \displaystyle -(6+2i) so we add the square of this expression to both sides

Expanding the square on the right-hand side \displaystyle \ (-(6+2i))^2=36+24i+4i^2=32+24i\ and completing the square on the left-hand side gives

After a taking roots, we have that \displaystyle \ z-(6+2i)=\pm 6\ and therefore the solutions are \displaystyle z=12+2i and \displaystyle z=2i.

Example 14

Complete the square in the expression \displaystyle \ z^2+(2-4i)z+1-3i\,.

Add and subtract the term \displaystyle \bigl(\frac{1}{2}(2-4i)\bigr)^2=(1-2i)^2=-3-4i\,,

Example 15

Calculate \displaystyle \ \sqrt{-3-4i}\,.

Put \displaystyle \ x+iy=\sqrt{-3-4i}\ where \displaystyle x and \displaystyle y are real numbers. Squaring both sides gives

which leads to the system of equations

From the second equation, we can solve for \displaystyle \ y=-4/(2x) = -2/x\ and put it into the first equation to get

This is a quadratic equation in \displaystyle x^2 which can be seen more easily by putting \displaystyle t=x^2,

The solutions are \displaystyle t = 1 and \displaystyle t = -4. The latter solution must be rejected, as \displaystyle x and \displaystyle y have been assumed to be real numbers, and thus \displaystyle x^2=-4 cannot be true. We get \displaystyle x=\pm\sqrt{1}, which gives us two possible solutions

• \displaystyle \ x=-1\ which gives \displaystyle \ y=-2/(-1)=2\,,
• \displaystyle \ x=1\ which gives \displaystyle \ y=-2/1=-2\,.

So we can conclude that

Example 16

1. Solve the equation \displaystyle \ z^2-2z+10=0\,.
   
   The formula for solutions to a quadratic equations (see example 3) gives that

   \displaystyle z= 1\pm \sqrt{1-10} = 1\pm \sqrt{-9}= 1\pm 3i\,\mbox{.}

2. Solve the equation \displaystyle \ z^2 + (4-2i)z -4i=0\,\mbox{.}
   
   Here, once again , the pq-formula may be used giving the solutions directly

   \displaystyle \begin{align*} z &= -2+i\pm\sqrt{\smash{(-2+i)^2+4i}\vphantom{i^2}} = -2+i\pm\sqrt{4-4i+i^{\,2}+4i}\\ &=-2+i\pm\sqrt{3} = -2\pm\sqrt{3}+i\,\mbox{.}\end{align*}

3. Solve the equation \displaystyle \ iz^2+(2+6i)z+2+11i=0\,\mbox{.}
   
   Division of both sides with \displaystyle i gives

   \displaystyle \begin{align*} z^2 + \frac{2+6i}{i}z +\frac{2+11i}{i} &= 0\,\mbox{,}\\ z^2+ (6-2i)z + 11-2i &= 0\,\mbox{.}\end{align*}

   Applying the pq- formula gives

   \displaystyle \begin{align*} z &= -3+i \pm \sqrt{\smash{(-3+i)^2 -(11-2i)}\vphantom{i^2}}\\ &= -3+i \pm \sqrt{-3-4i}\\ &= -3+i\pm(1-2i)\end{align*}

   where we used the resulting value of\displaystyle \ \sqrt{-3-4i}\ which we obtained in example 15. The solutions are therefore

   \displaystyle z=\biggl\{\begin{align*} &-2-i\,\mbox{,}\\ &-4+3i\,\mbox{.}\end{align*}