Tip:A–D to answerE for explanationV for videoS to reveal answer
Let z=1+i and z1=zˉ(1−z)+z11+izˉ. Then π12arg(z1) is equal to:
- A.
9
(Correct Answer) - B.
8
- C.
7
- D.
6
Explanation
Solution
Numerator: 1+i(1−i)=1+i+1=2+i
Denominator: (1−i)(−i)+21−i=−i−1+0.5−0.5i=−0.5−1.5i
z1=−0.5−1.5i2+i=−(1+3i)4+2i=−(1+9)(4+2i)(1−3i)=−1010−10i=−1+i
arg(z1)=π−4π=43π
π12(43π)=9
Answer: 9
Explanation
Solution
Numerator: 1+i(1−i)=1+i+1=2+i
Denominator: (1−i)(−i)+21−i=−i−1+0.5−0.5i=−0.5−1.5i
z1=−0.5−1.5i2+i=−(1+3i)4+2i=−(1+9)(4+2i)(1−3i)=−1010−10i=−1+i
arg(z1)=π−4π=43π
π12(43π)=9
Answer: 9
