Let z be a complex number such that |z+2|=1 a | vedclass

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(D) Given $|z+2|=1$, we can write $z+2 = \cos 	heta + i \sin 	heta$ for some $	heta \in [0, 2\pi)$.Then $\frac{1}{z+2} = \cos 	heta - i \sin 	het

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$\frac{2 \sqrt{6}}{5}$

Vedclass · Answer

(D) Given $|z+2|=1$, we can write $z+2 = \cos 	heta + i \sin 	heta$ for some $	heta \in [0, 2\pi)$.Then $\frac{1}{z+2} = \cos 	heta - i \sin 	heta$.We have $\frac{z+1}{z+2} = \frac{z+2-1}{z+2} = 1 - \frac{1}{z+2} = 1 - (\cos 	heta - i \sin 	heta) = (1 - \cos 	heta) + i \sin 	heta$.Given $\operatorname{Im}\left(\frac{z+1}{z+2}ight) = \frac{1}{5}$, we get $\sin 	heta = \frac{1}{5}$.Since $\cos^2 	heta = 1 - \sin^2 	heta = 1 - \frac{1}{25} = \frac{24}{25}$, we have $\cos 	heta = \pm \frac{2 \sqrt{6}}{5}$.Now, $\overline{z+2} = \overline{\cos 	heta + i \sin 	heta} = \cos 	heta - i \sin 	heta$.Thus, $\operatorname{Re}(\overline{z+2}) = \cos 	heta$.Therefore, $|\operatorname{Re}(\overline{z+2})| = |\cos 	heta| = \left| \pm \frac{2 \sqrt{6}}{5} ight| = \frac{2 \sqrt{6}}{5}$.

List-$I$	List-$II$
$(P)$ $\|z\|^2$ is equal to	$(1)$ $12$
$(Q)$ $\|z-\bar{z}\|^2$ is equal to	$(2)$ $4$
$(R)$ $\|z\|^2+\|z+\bar{z}\|^2$ is equal to	$(3)$ $8$
$(S)$ $\|z+1\|^2$ is equal to	$(4)$ $10$
	$(5)$ $7$

Let $z$ be a complex number such that $|z+2|=1$ and $\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$. Then the value of $|\operatorname{Re}(\overline{z+2})|$ is:

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