Mathematics - Complex Number Question with Solution | TestHub

P: If $z^3=1$ and $z\ne 1$, then $z$ is $\omega$ or ${\omega }^2$. In both cases, $z+\bar{z}=\omega +{\omega }^2=-1$. Thus, P matches 3.

Q: The non-real roots of $z^5-1=0$ are $e^{i2k\pi /5}$ for $k=1,2,3,4$. The condition $z+1/z=2\cos (2\pi /5)$ implies $2\cos (\arg (z))=2\cos (2\pi /5)$. This means $\arg (z)=\pm 2\pi /5$. So $z=e^{i2\pi /5}$ or $z=e^{-i2\pi /5}$. The option $z=\cos (2\pi /5)+i\sin (2\pi /5)$ is $e^{i2\pi /5}$. Thus, Q matches 2.

R: Let $z=x+iy$. The condition $|z-1|=1$ implies $(x-1{)}^2+y^2=1$. The condition $\arg (z)=\pi /4$ implies $y=x$ and $x>0$. Substituting $y=x$ into the first equation: $(x-1{)}^2+x^2=1⟹x^2-2x+1+x^2=1⟹2x^2-2x=0⟹2x(x-1)=0$. Since $x>0$, we must have $x=1$. Thus $z=1+i$. For $z=1+i$, $\text{Re}(z)=1$ and $\text{Im}(z)=1$, so $\text{Re}(z)=\text{Im}(z)$. Thus, R matches 4.

S: $z^3=-1=e^{i\pi }$. The cube roots of $-1$ are $e^{i(\pi +2k\pi )/3}$ for $k=0,1,2$. These are $e^{i\pi /3},e^{i\pi },e^{i5\pi /3}$. We need $\text{Re}(z)>0$ and $\text{Im}(z)>0$, which means $z$ must be in the first quadrant. Only $z=e^{i\pi /3}=\cos (\pi /3)+i\sin (\pi /3)=1/2+i\sqrt{3}/2$ satisfies this condition. For this $z$, $\text{Re}(z)=1/2$. Thus, S matches 1.