Let z be a complex number such that | z + 2 | = 1 and Im ( z + 1 z + 2 ) = 1 5 . Then the value of | R e ( z + 2 ¯ ) | is [2024]

Question

Let $z$ be a complex number such that $| z + 2 | = 1$ and $Im$ $(\frac{z + 1}{z + 2})$ $= \frac{1}{5} .$ Then the value of $| R e (\bar{z + 2}) |$ is [2024]

Smart Achievers · Accepted Answer

(1)

Let $| z + 2 | = 1$

$Im$ ...(i)

Also, $Im$ $(\frac{(x + 1) + i y}{(x + 2) + i y})$ $= \frac{1}{5}$

$\Rightarrow Im$ $(\frac{[(x + 1) + i y] [(x + 2) - i y]}{{(x + 2)}^{2} + y^{2}})$ $= \frac{1}{5}$

$\Rightarrow \frac{(x + 2) y - (x + 1) y}{{(x + 2)}^{2} + y^{2}} = \frac{1}{5}$

$\Rightarrow \frac{y}{1} = \frac{1}{5}$ [Using (i)]

$\Rightarrow {(x + 2)}^{2} + \frac{1}{25} = 1 \Rightarrow {(x + 2)}^{2} = 1 - \frac{1}{25}$

$\Rightarrow (x + 2) = \pm \frac{2 \sqrt{6}}{5}$

$∴$ $| Re (\bar{z + 2}) |$ $=$ $| Re[(x + 2) - i y)] |$ $=$ $| (x + 2) |$ $= \frac{2 \sqrt{6}}{5} .$

Solution

Q.
Let $z$ be a complex number such that $| z + 2 | = 1$ and $Im$ $(\frac{z + 1}{z + 2})$ $= \frac{1}{5} .$ Then the value of $| R e (\bar{z + 2}) |$ is [2024]

1 $\frac{2 \sqrt{6}}{5}$

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

3 $\frac{1 + \sqrt{6}}{5}$

4 $\frac{24}{5}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let z be a complex number such that |z+2|=1 and Im(z+1z+2)=15. Then the value of |Re(z+2¯)| is [2024]

1 265

2 65

3 1+65

4 245