1. ## Equation and parameter

Solve in C :
t is real parameter
Find the modulus and argument of this solutions .

2. Hint: let $z=r ( cos \theta+i sin \theta)$

3. $\Delta=b^2-4ac=4-\frac{4}{\cos^2t}=\frac{4(\cos^2t)}{\cos^2t}=-4\tan^2t$

$z_{1,2}=\frac{2\pm 2i\tan t}{2}=1\pm i\tan t$

$z_1=\frac{1}{\cos t}\left(\cos t+i\sin t\right)$

$z_2=\frac{1}{\cos t}\left(\cos t-i\sin t\right)=\frac{1}{\cos t}\left(\cos(2\pi-t)+i\sin(2\pi-t)\right)$

If $t\in\left[\left.0,\frac{\pi}{2}\right)\right.\Rightarrow|z_1 |=|z_2|=\frac{1}{\cos t}$

If $t\in\left(\left.\frac{\pi}{2},\pi\right]\right.\Rightarrow|z_1|=|z_2|=-\frac{1}{\cos t}$

$\arg z_1=t, \ \arg z_2=2\pi-t$

4. Originally Posted by red_dog
$\Delta=b^2-4ac=4-\frac{4}{\cos^2t}=\frac{4(\cos^2t)}{\cos^2t}=-4\tan^2t$

$z_{1,2}=\frac{2\pm 2i\tan t}{2}=1\pm i\tan t$

$z_1=\frac{1}{\cos t}\left(\cos t+i\sin t\right)$

$z_2=\frac{1}{\cos t}\left(\cos t-i\sin t\right)=\frac{1}{\cos t}\left(\cos(2\pi-t)+i\sin(2\pi-t)\right)$

If $t\in\left[\left.0,\frac{\pi}{2}\right)\right.\Rightarrow|z_1 |=|z_2|=\frac{1}{\cos t}$

If $t\in\left(\left.0,\frac{\pi}{2}\right]\right.\Rightarrow|z_1|=|z_2|=-\frac{1}{\cos t}$

$\arg z_1=t, \ \arg z_2=2\pi-t$
Hello THANK YOU but I'have this remak :

5. Originally Posted by dhiab
Hello THANK YOU but I'have this remak :

In my country $\arg z$ is called the reduced argument and $\arg z\in[0,2\pi]$

$Arg (z)$ means all the arguments and $Arg (z)=\arg z+2k\pi$

In this case $0\leq t\leq \pi, \ t\neq\frac{\pi}{2}$

$-\pi\leq-t\leq 0$ and $-t$ is not the reduced argument. Then we have to reduce it by adding some periods of sine and cosine. It is sufficiently to add one period. So $\arg z=2\pi-t$