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  • For two non-zero complex numbers <img alt="\mathrm{z_1 \: \: and\: \: z_2,}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bz_1%20%5C%3A%20%5C%3A%20and%5C%3A%20%5C%3A%20z_2%2C%7D"

For two non-zero complex numbers \mathrm{z_1 \: \: and\: \: z_2,} if \mathrm{\operatorname{Re}\left(z_1 z_2\right)=0\: \: and \: \: \operatorname{Re}\left(z_1+z_2\right)=0}, then which of the following are possible?
\mathrm{A. \operatorname{Im}\left(z_1\right)>0 \: and\: \operatorname{Im}\left(z_2\right)>0}
\mathrm{B. \operatorname{Im}\left(z_1\right)<0\: and \: \operatorname{Im}\left(z_2\right)>0}
\mathrm{C. \operatorname{Im}\left(z_1\right)>0 \: and\: \operatorname{Im}\left(z_2\right)<0}
\mathrm{D. \operatorname{Im}\left(z_1\right)<0 \: and\: \operatorname{Im}\left(z_2\right)<0}
Choose the correct answer from the options given below:


 

Option: 1

\mathrm{B\: and \: D}
 


Option: 2

\mathrm{A \: and \: B}
 


Option: 3

\mathrm{B\: and \: C}
 


Option: 4

\mathrm{A \: and \: C}


Answers (1)

best_answer

\mathrm{\operatorname{Re}\left(z_1 z_2\right)=0 \: and \: \operatorname{Re}\left(z_1+z_2\right)=0}
Let \mathrm{z_1=a_1+i b_1\: and \: z_2=a_2+i b_2}
\begin{aligned} & \mathrm{z}_1 \mathrm{z}_2=\left(\mathrm{a}_1 \mathrm{a}_2-\mathrm{b}_1 \mathrm{~b}_2\right)+\mathrm{i}\left(\mathrm{a}_1 \mathrm{~b}_2+\mathrm{b}_1 \mathrm{a}_2\right) \\ & \because \operatorname{Re}\left(\mathrm{z}_1 \mathrm{z}_2\right)=\mathrm{a}_1 \mathrm{a}_2-\mathrm{b}_1 \mathrm{~b}_2=0 \\ & \therefore \mathrm{a}_1 \mathrm{a}_2=\mathrm{b}_1 \mathrm{~b}_2 \ldots \ldots \ldots(1) \end{aligned}
\mathrm{and \operatorname{Re}\left(\mathrm{z}_1+\mathrm{z}_2\right)=0 \Rightarrow \mathrm{a}_1+\mathrm{a}_2=0}

\Rightarrow \mathrm{a_{2}=-a_{1}}................(2)

from (1) and (2)

\Rightarrow \mathrm{b_{1}b_{2}=-a^{2}_{1}< 0}

Product of \mathrm{b_{1}b_{2}} is Negative

\therefore \mathrm{Im\left ( z_{1} \right )\: and \: Im\left ( z_{2} \right )} are also of opposite sign

 

Posted by

Nehul

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