Exercises: complex numbers – Serlo

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Real and imaginary part[Bearbeiten]

Exercise (Determining imaginary and real part)

Determine the real part and the imaginary part of the following complex numbers:

Solution (Determining imaginary and real part)

Solution sub-exercise 1:

There is

So and .

Solution sub-exercise 2:

There is

So and .

Solution sub-exercise 3:

There is

So and .

Plane of complex numbers[Bearbeiten]

Exercise

Sketch the following sets on the plane of complex numbers:

Solution

Solution sub-exercise 1:

The set describes all complex numbers which fulfil the condition . In the complex plane this corresponds exactly to the numbers , whose distance from the number is at most . So describes a circle around with radius . The boundary of the circle is included in the set:

The set M1

Solution sub-exercise 2:

For all elements of the set , should apply. So their distance from the origin should be at least , but at most . Thus the set describes a circular ring between the circles with the radii and , including the boundary lines:

The set M2

Solution sub-exercise 3:

The set describes a circle around the origin with radius :

First part of the set M3

The second part of the set describes all complex numbers whose real part is larger than their imaginary part. On the bisector of the first and third quadrant ("diagonal line") we have are all numbers whose real and imaginary part are equal. To the right of this line, there are all complex numbers with a real part greater than the imaginary part:

Second part of the set M3

The intersection of both sets yields:

Intersection of the sets

With this you can sketch the set . Note that one of the boundary lines is only dashed. That means, the numbers directly on the origin line do not belong to the set , since the real part of each element must be strictly bigger than its imaginary part:

Menge M3

Polar representation[Bearbeiten]

Exercise (Transform to polar representation)

Compute the complex numbers and convert them into polar representation:

Solution (Transform to polar representation)

Solution sub-exercise 1:

The absolute value is . Since and there is . Thus . This formula appears frequently, so you might want to learn it by heart.

Solution sub-exercise 2:

We have the absolute value . With and the number lies in the fourth quadrant. The formula for this is . So there is .

Solution sub-exercise 3:

You get the polar representation of by first complexly conjugating the polar representation of and then multiplying it by . There is:

Under complex conjugation, is replaced by in the exponent in the polar representation. Thus .

Solution sub-exercise 4:

By multiplying it out, we get . So . With and the number lies in the second quadrant. For the angle there is . This corresponds in degrees to an angle of . Altogether we get the result .

Solution sub-exercise 5:

First, we determine the polar representation of the three factors of :

Then, there is

Solution sub-exercise 6:

First we determine the absolute value and the angle of . There is and . Now we can determine the polar representation of :