Search for: Let z=32+i25+32−i25 .If R(z) and I(z) respectively denote the real and imaginary parts of z, thenLet z=32+i25+32−i25 .If R(z) and I(z) respectively denote the real and imaginary parts of z, thenAR(z)>0 and I(z)>0 BI(z)=0 CR(z)<0 and I(z)=0DR(z)=−3 Register to Get Free Mock Test and Study Material +91 Verify OTP Code (required) I agree to the terms and conditions and privacy policy. Solution:We have z=32+i25+32−i23=cosπ6+isinπ65+cosπ6−isinπ65=cos5π6+isin5π6+cos5π6−isin5π6=2cos5π6<0Here, I(z)>0 and R(z)>0. Post navigationPrevious: lf x = 33n, n is a positive integral value, then what is the probability that x will have 3 at its units place?Next: The value of cosπ22⋅cosπ23⋅⋯⋅cosπ210⋅sinπ210 is Related content JEE Main 2023 Question Papers with Solutions JEE Main 2024 Syllabus Best Books for JEE Main 2024 JEE Advanced 2024: Exam date, Syllabus, Eligibility Criteria JEE Main 2024: Exam dates, Syllabus, Eligibility Criteria JEE 2024: Exam Date, Syllabus, Eligibility Criteria NCERT Solutions For Class 6 Maths Data Handling Exercise 9.3 JEE Crash Course – JEE Crash Course 2023 NEET Crash Course – NEET Crash Course 2023 JEE Advanced Crash Course – JEE Advanced Crash Course 2023