If z₁=x₁+i y₁, \;\; z₂=x₂+i y₂, z₃=x₁+{i x₂}{2}, z₄=2 y₁+i y₂ are complex numbers such that |z₁|=1,|z₂|=2 and {Re}(z₁ z₂)=0, then

Correct Answer is : ∣z3∣=1,∣z4∣=2,Re⁡(z3z4)=0|z₃|=1,|z₄|=2,{Re}(z₃z₄)=0∣z3∣=1,∣z4∣=2,Re(z3z4)=0

|z₃|=1,|z₄|=2,{Im}(z₃z₄)=0

|z₃|=2,|z₄|=1,{Re}(z₃z₄)=0

|z₃|=1,|z₄|=2,{Re}(z₃z₄)=0

|z₃|=2,|z₄|=1,{Re}(z₁z₃)={Im}(z₂z₄)=0

Correct Answer is : ∣z3∣=1,∣z4∣=2,Re⁡(z3z4)=0|z₃|=1,|z₄|=2,{Re}(z₃z₄)=0∣z3∣=1,∣z4∣=2,Re(z3z4)=0

Test Index

TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 7 of 160

Marks: +1, -0

z_{1}=x_{1}+i y_{1}, \;\; z_{2}=x_{2}+i y_{2}, z_{3}=x_{1}+\frac{i x_{2}}{2}, z_{4}=2 y_{1}+i y_{2}

|z_{1}|=1,|z_{2}|=2

\operatorname{Re}(z_{1} z_{2})=0

, then

Solution:

We have,

z_{1} \;\; =x_{1}+i y_{1}

z_{2} \;\; =x_{2}+i y_{2}

z_{3} \;\; =x_{1}+\frac{i x_{2}}{2}

z_{4} \;\; =2 y_{1}+i y_{2}

|z_{1}| \;\; =1,|z_{2}|=2 \;\text{ and }\; \operatorname{Re}(z_{1} z_{2})=0

\;\text{ Let }\; z_{1}=e^{i \alpha} \;\text{ and }\; z_{2}=2 e^{i \beta}

z_{1} z_{2} \;\; =2 e^{i(\alpha+\beta)}

z_{1} z_{2} \;\; =2(\cos (\alpha+\beta)+i \sin (\alpha+\beta))

\operatorname{Re}(z_{1}+z_{2})=2 \& \cos (\alpha+\beta)=0

\therefore \;\; \alpha+\beta=\frac{\pi}{2}

\beta=\frac{\pi}{2}-\alpha

\therefore \;\; z_{1} \;\; =\cos \alpha+i \sin \alpha

z_{2} \;\; =2(\cos \beta+i \sin \beta)

z_{2} \;\; =2(\sin \alpha+i \cos \alpha)

\therefore \;\; z_{3} \;\; =\cos \alpha+i\left(\frac{2 \sin \alpha}{2}\right)

z_{3} \;\; =\cos \alpha+i \sin \alpha

|z_{3}| \;\; =1

z_{4} \;\; =2(\sin \alpha+i \cos \alpha)

|z_{4}| \;\; =2

z_{3} z_{4} \;\; =2(\cos \alpha+i \sin \alpha)(\sin \alpha+i \cos \alpha)

z_{3} z_{4} \;\; =2[(\cos \alpha \sin \alpha-\cos \alpha \sin \alpha)

+i(\sin ^{2} \alpha+\cos ^{2} \alpha)]

z_{3} z_{4} \;\; =2 i

\operatorname{Re}(z_{3} z_{4}) \;\; =0

Go to Question:

Prev Question

Next Question