A nonzero complex number is \(z=re^{i\theta}\) with \(r=\abs{z}\), and its \(n\)th roots are \(r^{1/n}e^{i(\theta+2\pi k)/n}\) for \(k=0,\ldots,n-1\): one root multiplied by the \(n\)th roots of unity. The primitive \(n\)th roots of unity generate the cyclotomic field \(\QQ(e^{2\pi i/n})\), of degree \(\varphi(n)\) over \(\QQ\). Cauchy’s integral formula: if \(f\) is holomorphic on an open set containing the closed disc \(\abs{z-z_0}\le r\), then \(f(a)=\frac{1}{2\pi i}\oint_{\abs{z-z_0}=r}\frac{f(z)}{z-a}\,dz\) for \(\abs{a-z_0}<r\).
Cube roots of $8$ in $\mathbb{C}$ (Tag P-3373K)
Find all cube roots of \(8\) in the complex plane. Write your answers in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and justify your answer.
Solution
Write \(8=8e^{2\pi i k}\) for \(k\in\mathbb Z\). Its cube roots are \[ 2e^{2\pi i k/3},\qquad k=0,1,2. \] Thus the three roots are \[ 2, \qquad 2\left(-\frac12+\frac{\sqrt3}{2}i\right)=-1+\sqrt3\,i, \qquad 2\left(-\frac12-\frac{\sqrt3}{2}i\right)=-1-\sqrt3\,i. \] Cubing each gives \(8\), and a degree-\(3\) polynomial has at most three roots, so this list is complete.
The cube roots of $2-2i$ (Tag P-OQW4B)
Find all cube roots of \(2-2i\) and express them in the standard form \(a + bi\).
Solution
<1>1. Write \(2-2i\) in polar form as \[ 2-2i=2\sqrt2\,e^{-i\pi/4}. \]
<1>2. Every cube root therefore has the form \[ z_k=\sqrt2\,e^{i(-\pi/12+2\pi k/3)},\qquad k=0,1,2. \]
The modulus of a cube root is \((2\sqrt2)^{1/3}=\sqrt2\), and its argument is one third of \(-\pi/4+2\pi k\).
<1>3. For \(k=0\), \[ z_0=\frac{1+\sqrt3}{2}-\frac{\sqrt3-1}{2}i. \]
Using \(\cos 15^\circ=(\sqrt6+\sqrt2)/4\) and \(\sin 15^\circ=(\sqrt6-\sqrt2)/4\), \[ \sqrt2(\cos15^\circ-i\sin15^\circ) =\frac{1+\sqrt3}{2}-\frac{\sqrt3-1}{2}i. \]
<1>4. For \(k=1\), \[ z_1=\frac{1-\sqrt3}{2}+\frac{1+\sqrt3}{2}i. \]
Here the argument is \(105^\circ\), so \[ \sqrt2(\cos105^\circ+i\sin105^\circ) =\frac{1-\sqrt3}{2}+\frac{1+\sqrt3}{2}i. \]
<1>5. For \(k=2\), \[ z_2=-1-i. \]
Here the argument is \(225^\circ\), and \[ \sqrt2(\cos225^\circ+i\sin225^\circ)=-1-i. \]
<1>6. These are all the cube roots of \(2-2i\).
The equation \(z^3=2-2i\) has degree \(3\), and <1>2 gives three distinct roots.
Cube and square roots of $-2+2\sqrt{3}i$ (Tag P-Q5LN7)
Let \(\alpha\) be the complex number \(-2+2\sqrt{3}i\). Express \(\alpha^3\) and the two square roots of \(\alpha\) in the standard form \(a+bi\).
Solution
The number \[ \alpha=-2+2\sqrt3\,i \] has modulus \(4\) and argument \(2\pi/3\), so \[ \alpha=4e^{2\pi i/3}. \] Hence \[ \alpha^3=4^3e^{2\pi i}=64. \] Its square roots are \[ \pm 2e^{\pi i/3} =\pm(1+\sqrt3\,i). \] Thus the two square roots are \[ 1+\sqrt3\,i,\qquad -1-\sqrt3\,i. \]
Cauchy integral formula for $\oint_\gamma z^2/(z-i)\,dz$ on a circle about $i$ (Tag P-FBY23)
- State the Cauchy integral formula and use it to evaluate the complex line integral: \(\oint_\gamma z^2/(z-i)\, dz\), where \(\gamma\) is a circle centered at \(i\) and oriented counterclockwise.
- Use a parametrization of the path \(\gamma\) to express the line integral explicitly in terms of real integrals, with limits of integration. [You do not have to evaluate these real integrals.]
Solution
Cauchy’s integral formula states that if \(f\) is holomorphic on and inside a positively oriented simple closed contour \(\gamma\), and \(a\) lies inside \(\gamma\), then \[ \oint_\gamma \frac{f(z)}{z-a}\,dz=2\pi i\,f(a). \] Taking \(f(z)=z^2\) and \(a=i\) gives \[ \oint_\gamma \frac{z^2}{z-i}\,dz =2\pi i\,i^2 =-2\pi i. \]
Let the circle have radius \(R>0\). A counterclockwise parametrization is \[ z(t)=i+Re^{it},\qquad 0\le t\le2\pi, \] with \[ dz=iRe^{it}\,dt. \] Therefore \[ \oint_\gamma \frac{z^2}{z-i}\,dz =\int_0^{2\pi} i\bigl(i+Re^{it}\bigr)^2\,dt. \] Writing \(e^{it}=\cos t+i\sin t\), this becomes \[ \int_0^{2\pi} \Bigl[-2R\cos t\,(1+R\sin t) +i\bigl(R^2\cos^2t-(1+R\sin t)^2\bigr)\Bigr]dt, \] which is an explicit expression in real integrals with limits of integration.