Complex numbers exercises

complex-numbers

 

We offer exercises corrected on complex numbers. Our goal is to show students how to calculate on the complex plane; how to solve algebraic equations on the complex plane; how to calculate the values ​​of trigonometric functions at different numbers.

Exercise: Rewrite the following complex numbers in the form $x + iy$: \begin{align*} z_1=\frac{1}{3+i},\quad z_2=\frac{1}{i^3},\qquad z_3=i(i+1)(i-2)^2. \end{align*}

Solution: We can write \begin{align*} z_1&=\frac{1}{3+i}\times \frac{3-i}{3-i}\cr &=\frac{3-i}{9+1}\cr &= \frac{3}{10}-\frac{1}{10}i. \end{align*}

Observe that $i^3=i^2 i=-i$. Then \begin{align*} z_2=\frac{1}{-i}=i. \end{align*}

\begin{align*} z_3&=i(i+1)(i-2)^2\cr&= (-1+i)(-1+2i+4)\cr&= 1-2i-4-i-2+4i\cr &=-5+i. \end{align*}

Exercise: rewrite in the polar and the exponential polar form: \begin{align*} z_1=-1 – i \sqrt{3},\quad z_2= 3 + 3i,\qquad z_3=-\frac{4}{3}i \end{align*}

Solution: The modulus of $z_1$ is \begin{align*} |z_1|=\sqrt{(-1)^2+(-\sqrt{3})^2}=\sqrt{4}=2. \end{align*} Let $\theta_1$ be the argument of $z_1$. We then have \begin{align*} \cos(\theta_1)&=\frac{{\rm Re}(z_1)}{|z_1|}=\frac{-1}{2},\cr \sin(\theta_1)&=\frac{{\rm Im}(z_1)}{|z_1|}=\frac{-\sqrt{3}}{2}. \end{align*} This shows that \begin{align*} \theta_1=\frac{4\pi}{3}+2k\pi,\quad k\in\mathbb{Z}. \end{align*} Thus \begin{align*} z_1=|z_1|e^{i\theta}=2 e^{i\frac{4\pi}{3}}. \end{align*} We need to give a representation of $z_1$ with a principal argument (i.e. belonging to $]-\pi,\pi]$). We can write $\frac{4\pi}{3}=\frac{-2\pi}{3}+2\pi$. Then \begin{align*} z_1==2 e^{i\frac{-2\pi}{3}}=2\left( \cos(\frac{-2\pi}{3})+i\sin(\frac{-2\pi}{3})\right). \end{align*}

We have $z_2=3(1+i)$. Then $|z_2|=3\sqrt{1^2+1^2}=3\sqrt{2}$. Let $\theta_2$ be an argument of $z_2$. We have \begin{align*} \cos(\theta_2)=\frac{3}{3\sqrt{2}}=\frac{\sqrt{2}}{2}. \end{align*} Similarly, \begin{align*} \sin(\theta_2)=\frac{\sqrt{2}}{2}. \end{align*} We then obtain \begin{align*} \theta_2=\frac{\pi}{4}+2k\pi,\quad k\in\mathbb{Z}. \end{align*} Hence \begin{align*} z_2&=|z_2|e^{i\theta_2}=3\sqrt{2} e^{i\frac{\pi}{4}}\cr &= 3\sqrt{2}\left(\cos(\frac{\pi}{4})+\sin(\frac{\pi}{4})\right). \end{align*}

Observe that $i=e^{i\frac{\pi}{2}}$. Then \begin{align*} z_3=-\frac{4}{3}i=-\frac{4}{3}e^{i\frac{\pi}{2}}. \end{align*} But this is not the good form as $-\frac{4}{3} < 0$. To overcome this obstacle we write $(-1)=e^{i\pi}$, so that \begin{align*} z_3&=\frac{4}{3}e^{i\pi} e^{i\frac{\pi}{2}}\cr &= \frac{4}{3} e^{i\frac{3\pi}{2}} \end{align*} We need a representation with a principal argument. We then write $e^{i\frac{3\pi}{2}}=e^{i\frac{4\pi-\pi}{2}}$. Thus \begin{align*} z_3=\frac{4}{3} e^{i\frac{-\pi}{2}}=\frac{4}{3} \left(\cos(\frac{-\pi}{2})+i\sin(\frac{-\pi}{2})\right). \end{align*}

Exercise: Solve the equation $z^2=1+i$.

Solution: We recall that if $\alpha= r e^{i\theta}$ (here $r>0$ and $\theta$ is a principal argument of $\alpha$), then the complex equation $z^2=\alpha$ has two roots: \begin{align*}z_1&=\sqrt{r}e^{i\frac{\theta}{2}}\;\text{and,}\cr z_2&=\sqrt{r}e^{i\left(\frac{\theta}{2}+\pi\right)}\end{align*} On the other hand, if we write $z$ and $\alpha$ is the algebraic form $z=x+iy$ and $\alpha=a+i b$ then using the fact that $z^2=x^2-y^2+i (2xy),$ we obatin \begin{align*} x^2-y^2=a,\quad 2xy=b. \end{align*} It suffices then to solve the above system to obtain $z$. After a small effort we obtain \begin{align*} |x|=\sqrt{\frac{\sqrt{a^2+b^2}+a}{2}},\quad |y|=\sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}. \end{align*} (to determine $x$ and $y$ we shall remark that the sign of the product $xy$ is the same as the sign of $b$, this is very important!!!).

Let us now sole the equation $z^2=1+i$ here we have $\alpha=1+i$ and then $a=1$ and $b=1$. Moreover, $\alpha=\sqrt{2}e^{i\frac{\pi}{4}}$, so that the principal argument of $\alpha$ is $\theta=\frac{\pi}{4}$. We have the two roots \begin{align*} z_1=\sqrt{\sqrt{2}}e^{i\frac{\pi}{8}},\quad z_2=\sqrt{\sqrt{2}}e^{i\frac{9\pi}{8}}=-z_1. \end{align*}

Si $z=x+iy$ then we have \begin{align*} \begin{cases} x^2-y^2=1\cr x^2+y^2=\sqrt{2}\cr 2xy=1 \end{cases} \end{align*} Thus \begin{align*} x^2=\frac{1+\sqrt{2}}{2},\quad y^2=\frac{-1+\sqrt{2}}{2}. \end{align*} As $x$ and $y$ have the same sign, we then have \begin{align*} z_1= \sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}},\quad z_2=-z_1. \end{align*}

From the exponential representation of the roots, we immediately deduce that \begin{align*} \cos(\frac{\pi}{8})&=\frac{\sqrt{1+\sqrt{2}}}{\sqrt{2\sqrt{2}}}\cr \sin(\frac{\pi}{8})&=\frac{\sqrt{-1+\sqrt{2}}}{\sqrt{2\sqrt{2}}} \end{align*}

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