Exercises · Numbers and logic · with worked solutions · PDF
Exercise 1
Write the following complex numbers \(z\) in algebraic form, identifying their real part and imaginary part. Then, for each of them, find the modulus, the conjugate and the reciprocal.
Representing the solutions in the Cartesian plane, we obtain a closed half-plane.
Exercise 3
Determine all the complex numbers \(z\) that satisfy:
\[
|z-(3+i)|\leq2
\]
Solution
This inequality is satisfied by those complex numbers \(z=x+iy\) which, in the Gauss plane (complex plane), are represented by the points \(P(x, y)\) whose distance from the point \((3, 1)\) is less than or equal to 2.
That is, all and only the points of the closed disk with center \(C(3, 1)\) and radius 2 (careful: it is not the circle, but the disk, i.e., the interior together with the boundary). The circle has equation:
\[
\gamma: \quad (x-3)^2+(y-1)^2=4
\]
while the closed disk has equation:
\[
(x-3)^2+(y-1)^2\leq4
\]
Alternatively, the inequality can be handled through the substitution \(z=x+iy\), obtaining
\[
|x+iy-3-i|\leq2
\]
in which, factoring out the imaginary unit,
\[
|x-3+i(y-1)|\leq2
\]
Applying the definition of the modulus of a complex number, we have
\[
\sqrt{(x-3)^2+(y-1)^2}\leq2
\]
Since we can square both non-negative sides, we finally obtain:
\[
(x-3)^2+(y-1)^2\leq4
\]
Exercise 4
Determine all the complex numbers \(z\) that satisfy:
Representing the solutions in the Cartesian plane, we obtain the part of the plane common to the closed disk with center at the origin and radius \(\sqrt{2}\) and to the closed half-plane to the left of the vertical line \(x=1\).
which, in the Gauss plane, can be represented as two points on the real axis. In conclusion, \(w\) is purely imaginary for the two values of \(z\) found.
Exercise 12
Let \(z=\sqrt{3}-3i\). The number of solutions \(w\) with \(\Im(w)<0\) of the equation
We have \(|z|=\sqrt{12}\), \(z+\bar{z}=2\sqrt{3}\), hence
\[
w^{6}=12.
\]
The complex sixth roots of a positive real number are six, pairwise opposite, and two of them are real. Of the remaining four, two have positive imaginary part and two have negative imaginary part. The correct answer is (a). We nevertheless want to determine their values. Setting \(w=r(\cos\vartheta+i\sin\vartheta)\) and taking into account that \(|12|=12\), \({\rm arg} 12=2k\pi\), \(k\in \Z\), we obtain
We first solve the equation \(z^{6}=i\). Setting \(z=r(\cos\vartheta+i\sin\vartheta)\) and taking into account that \(|i|=1\), \({\rm arg}(i)=\pi/2+2k\pi\), \(k\in{\bf Z}\), we obtain
Solve the following equation in the complex variable \(z\).
\[
\left(\frac{z+i}{1+i}\right)^{3}=-8i
\]
Solution
Setting \(w=(z+i)/(1+i)\), we solve \(w^{3}=-8i\). Writing \(w=r(\cos\vartheta+i\sin\vartheta)\) and taking into account that \(|-8i|=8\), \({\rm arg}(-8i)=-\pi/2+2k\pi\), \(k\in \Z\), we obtain
Solve the following equation in the complex variable \(z\).
\[
\left(\frac{z-i}{i}\right)^{4}=-16
\]
Solution
From \(i^{4}=1\) we have \((z-i)^{4}=-16\). Setting \(w=z-i\), we solve \(w^{4}=-16\). Writing \(w=r(\cos\vartheta+i\sin\vartheta)\) and taking into account that \(|-16|=16\), \({\rm arg}(-16)=-\pi+2k\pi\), \(k\in \Z\), we obtain
Solve the following equation in the complex variable \(z\).
\[
z^5=-\bar{z}
\]
Solution
The equation can be seen as
\[
z^5=(-1)\bar{z}
\]
Setting \(z=r(\cos\vartheta+i\sin\vartheta)\) and taking into account that \(|\bar{z}|=r\), \({\rm arg}(\bar{z})=-\vartheta+2k\pi\), \(k\in \Z\), and recalling that \(-1=e^{i\pi}\), we obtain
The equation has real coefficients, hence the solutions come in conjugate pairs; in particular, another solution is \(z_{2}=-i\). It follows that the polynomial
\[
p(z)=z^{4}-2z^{3}+4z^{2}-2z+3
\]
is divisible by \((z-i)(z+i)\), i.e., by \(z^{2}+1\). Performing the division we obtain
\[
p(z)=(z^{2}+1)(z^{2}-2z+3)
\]
hence among the solutions we also have the roots of
\[
z^{2}-2z+3=0
\]
which are \(z_{3}=1-i\sqrt{2}\), \(z_{4}=1+i\sqrt{2}\).
Summarizing, the solutions of the given equation are