Let Z 1 and Z 2 be complex numbers such that z - Z 2 and - z 1- z 2- - If Re z 1-0 and Im z 2-0- then Z 1- Z 2- Z 1- Z 2 isA- oneB- real and positiveC- real and negativeD- purely imaginary

The correct option is D purely imaginaryLet z_1=a+ib, z_2=c+id |z_1|=|z_2| ⇒ a^2+b^2=c^2+d^2 ⇒ a^2+b^2 c^2 d^2=0 ...(1) z_1 + z_2z_1 z_2=(a+c)+i(b+d)(a c) ...

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Byju's Answer

Standard XII

Mathematics

Conjugate of a Complex Number

Let Z 1 and Z...

Question

Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq z_{2}$ and $| z_{1} | = | z_{2} | .$ If $R e (z_{1}) > 0$ and $I m (z_{2}) < 0,$ then $\frac{z_{1} + z_{2}}{z_{1} - z_{2}}$ is

one

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real and positive

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real and negative

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purely imaginary

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Solution

The correct option is D purely imaginary
Let $z_{1} = a + i b, z_{2} = c + i d$
$| z_{1} | = | z_{2} |$
$\Rightarrow a^{2} + b^{2} = c^{2} + d^{2}$
$\Rightarrow a^{2} + b^{2} - c^{2} - d^{2} = 0 . . . (1)$

$\frac{z_{1} + z_{2}}{z_{1} - z_{2}} = \frac{(a + c) + i (b + d)}{(a - c) + i (b - d)}$
$= [\frac{(a + c) + i (b + d)}{(a - c)^{2} + (b - d)^{2}}] \times [(a - c) - i (b - d)]$
$= \frac{2 i (a d - b c)}{(a - c)^{2} + (b - d)^{2}}$

$\Rightarrow \frac{z_{1} + z_{2}}{z_{1} - z_{2}}$ is purely imaginary.

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Let z1 and z2 be complex numbers such that z1≠z2 and |z1|=|z2|. If Re(z1)>0 and Im(z2)<0, then z1+z2z1−z2 is

The correct option is D purely imaginaryLet z1=a+ib, z2=c+id |z1|=|z2| ⇒a2+b2=c2+d2 ⇒a2+b2−c2−d2=0 ...(1) z1+z2z1−z2=(a+c)+i(b+d)(a−c)+i(b−d) =[(a+c)+i(b+d)(a−c)2+(b−d)2]×[(a−c)−i(b−d)] =2i(ad−bc)(a−c)2+(b−d)2 ⇒z1+z2z1−z2 is purely imaginary.

Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq z_{2}$ and $| z_{1} | = | z_{2} | .$ If $R e (z_{1}) > 0$ and $I m (z_{2}) < 0,$ then $\frac{z_{1} + z_{2}}{z_{1} - z_{2}}$ is