For two non-zero complex numbers z_1 and z_2 , if \operatornameRe(z_1 z_2)=0 and \operatornameRe(z_1+z_2)=0 , then which of the following are possible? A. \operatornameIm(z_1)\gt 0 and \operatornameIm(z_2) \gt 0 B. \operatornameIm(z_1) 0 C. \operatornameIm(z_1) \gt 0 and \operatornameIm(z_2) D. \operatornameIm(z_1) Choose the correct answer from the options given below :

Complex Numbers PYQ — JEE Main 2023 · 29 Jan · Shift 1 · Q31 with Solution

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Correct answer: THE MATHEMATICALLY CORRECT ANSWER IS: STATEMENTS C AND D ARE POSSIBLE., HENCE, IF THE PRINTED OPTIONS ARE ACCURATE, THE STORED ANSWER D (B AND C) IS INCORRECT., THERE IS LIKELY A FORMATTING/PRINTING ERROR IN THE QUESTION TEXT OR OPTION LIST.

Let

\qquad z_2=x_2+i y_2$$ with $x_1,x_2,y_1,y_2\in\mathbb R$ and $z_1,z_2\neq 0$. We are given: 1. $$\operatorname{Re}(z_1z_2)=0$$ 2. $$\operatorname{Re}(z_1+z_2)=0$$ We must determine which sign combinations of $\operatorname{Im}(z_1)=y_1$ and $\operatorname{Im}(z_2)=y_2$ are possible. --- ## 1. Use the condition on the real part of the sum Since $$z_1+z_2=(x_1+x_2)+i(y_1+y_2),$$ we get $$\operatorname{Re}(z_1+z_2)=x_1+x_2=0.$$ So, $$x_2=-x_1.$$ --- ## 2. Use the condition on the real part of the product Now, $$z_1z_2=(x_1+i y_1)(x_2+i y_2).$$ Its real part is $$\operatorname{Re}(z_1z_2)=x_1x_2-y_1y_2.$$ Given this is zero, $$x_1x_2-y_1y_2=0.$$ Using $x_2=-x_1$, $$x_1(-x_1)-y_1y_2=0$$ $$-x_1^2-y_1y_2=0$$ $$y_1y_2=-x_1^2.$$ Hence, $$y_1y_2\le 0.$$ So the imaginary parts must have **opposite signs** or one of them may be zero. They **cannot both be positive** and **cannot both be negative**. --- ## 3. Check each statement From the image text, the four statements correspond to: - **A:** $\operatorname{Im}(z_1)>0$ and $\operatorname{Im}(z_2)>0$ - **B:** $\operatorname{Im}(z_1)<0$ and $\operatorname{Im}(z_2)<0$ - **C:** $\operatorname{Im}(z_1)>0$ and $\operatorname{Im}(z_2)<0$ - **D:** $\operatorname{Im}(z_1)<0$ and $\operatorname{Im}(z_2)>0$ Now evaluate: ### Statement A If both are positive, then $y_1y_2>0$, but we need $y_1y_2=-x_1^2\le 0$. So **A is not possible**. ### Statement B If both are negative, then again $y_1y_2>0$, impossible. So **B is not possible**. ### Statement C If $y_1>0$ and $y_2<0$, then $y_1y_2<0$, which is possible. Example: $$z_1=1+i,\qquad z_2=-1+i?$$ This gives $y_2>0$, so not suitable. Instead take $$z_1=1+i,\qquad z_2=-1-i.$$ Then $$\operatorname{Re}(z_1+z_2)=\operatorname{Re}(0)=0,$$ and $$z_1z_2=(1+i)(-1-i)=-2i,$$ so $$\operatorname{Re}(z_1z_2)=0.$$ Thus **C is possible**. ### Statement D If $y_1<0$ and $y_2>0$, then $y_1y_2<0$, also possible. Example: $$z_1=1-i,\qquad z_2=-1+i.$$ Then $$\operatorname{Re}(z_1+z_2)=0,$$ and $$z_1z_2=(1-i)(-1+i)=2i,$$ so $$\operatorname{Re}(z_1z_2)=0.$$ Thus **D is possible**. --- ## 4. Final conclusion The possible cases are **C and D**. Among the given answer choices, that corresponds to **Option D** if the listed pairs are interpreted as opposite-sign imaginary parts. However, from the explicit statement labels above, the valid statements are **C and D**, not **B and C**. So the mathematical conclusion is: $$\boxed{\text{Statements C and D are possible}}$$ There appears to be a mismatch/typo in the option formatting of the question text.

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Let

\qquad z_2=x_2+i y_2$$ with $x_1,x_2,y_1,y_2\in\mathbb R$ and $z_1,z_2\neq 0$. We are given: 1. $$\operatorname{Re}(z_1z_2)=0$$ 2. $$\operatorname{Re}(z_1+z_2)=0$$ We must determine which sign combinations of $\operatorname{Im}(z_1)=y_1$ and $\operatorname{Im}(z_2)=y_2$ are possible. --- ## 1. Use the condition on the real part of the sum Since $$z_1+z_2=(x_1+x_2)+i(y_1+y_2),$$ we get $$\operatorname{Re}(z_1+z_2)=x_1+x_2=0.$$ So, $$x_2=-x_1.$$ --- ## 2. Use the condition on the real part of the product Now, $$z_1z_2=(x_1+i y_1)(x_2+i y_2).$$ Its real part is $$\operatorname{Re}(z_1z_2)=x_1x_2-y_1y_2.$$ Given this is zero, $$x_1x_2-y_1y_2=0.$$ Using $x_2=-x_1$, $$x_1(-x_1)-y_1y_2=0$$ $$-x_1^2-y_1y_2=0$$ $$y_1y_2=-x_1^2.$$ Hence, $$y_1y_2\le 0.$$ So the imaginary parts must have **opposite signs** or one of them may be zero. They **cannot both be positive** and **cannot both be negative**. --- ## 3. Check each statement From the image text, the four statements correspond to: - **A:** $\operatorname{Im}(z_1)>0$ and $\operatorname{Im}(z_2)>0$ - **B:** $\operatorname{Im}(z_1)<0$ and $\operatorname{Im}(z_2)<0$ - **C:** $\operatorname{Im}(z_1)>0$ and $\operatorname{Im}(z_2)<0$ - **D:** $\operatorname{Im}(z_1)<0$ and $\operatorname{Im}(z_2)>0$ Now evaluate: ### Statement A If both are positive, then $y_1y_2>0$, but we need $y_1y_2=-x_1^2\le 0$. So **A is not possible**. ### Statement B If both are negative, then again $y_1y_2>0$, impossible. So **B is not possible**. ### Statement C If $y_1>0$ and $y_2<0$, then $y_1y_2<0$, which is possible. Example: $$z_1=1+i,\qquad z_2=-1+i?$$ This gives $y_2>0$, so not suitable. Instead take $$z_1=1+i,\qquad z_2=-1-i.$$ Then $$\operatorname{Re}(z_1+z_2)=\operatorname{Re}(0)=0,$$ and $$z_1z_2=(1+i)(-1-i)=-2i,$$ so $$\operatorname{Re}(z_1z_2)=0.$$ Thus **C is possible**. ### Statement D If $y_1<0$ and $y_2>0$, then $y_1y_2<0$, also possible. Example: $$z_1=1-i,\qquad z_2=-1+i.$$ Then $$\operatorname{Re}(z_1+z_2)=0,$$ and $$z_1z_2=(1-i)(-1+i)=2i,$$ so $$\operatorname{Re}(z_1z_2)=0.$$ Thus **D is possible**. --- ## 4. Final conclusion The possible cases are **C and D**. Among the given answer choices, that corresponds to **Option D** if the listed pairs are interpreted as opposite-sign imaginary parts. However, from the explicit statement labels above, the valid statements are **C and D**, not **B and C**. So the mathematical conclusion is: $$\boxed{\text{Statements C and D are possible}}$$ There appears to be a mismatch/typo in the option formatting of the question text.

Complex Numbers question

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