A) 2 + 2i
B) - 2 - 2i
C) 1 + i
D) - 1 - i
Correct Answer: C
Solution :
Let P(2 + i) By rotation theorem \[\frac{z-(3+3i)}{3+i-(3+3i)}=\frac{2\sqrt{2}}{2}{{e}^{(-\pi /4)i}}\] \[\frac{z-3-3i}{-2i}=1-i\] \[z-3-3i=-2i-1\] \[z=1+i\]You need to login to perform this action.
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