3.3 Potenzen und Wurzeln

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 Moivres theorem gives

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Example 2

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

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and according to the Moivres theorem one gets

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

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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 Moivres Theorem

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

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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

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Matching the moduli and arguments on both sides gives

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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

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Example 6

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

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which demonstrates that Moivres theorm is actually identical to the well-known law of powers,

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Example 7

From the above definition, one can obtain the relationship

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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,

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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

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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

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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:
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   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:
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   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

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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

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

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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

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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

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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\,,

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Example 15

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

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

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which leads to the system of equations

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

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

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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

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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
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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 .
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3. Solve the equation \displaystyle \ iz^2+(2+6i)z+2+11i=0\,\mbox{.} Division of both sides with \displaystyle i gives
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   Applying the pq- formula gives
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   where we used the resulting value of\displaystyle \ \sqrt{-3-4i}\ which we obtained in example 15. The solutions are therefore
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