Q.
Let two non-zero distinct complex numbers z1 and z2 be represented respectively by points A and B in the complex plane O as the origin. Then OA is perpendicular to OB if an only if
1748
163
COMEDKCOMEDK 2005Complex Numbers and Quadratic Equations
Report Error
Solution:
OA and OB are perpendicular iff slope of OA× slope of OB=1(a1−0b1−0)⋅(a2−0b2−0)=−1 ⇒b1b2=−a1a2⇒b1b2+a1a2=0
This is possible in case of z1z2+z2z1=0
i.e., a1a2+a1b2i+b1a2i+b1b2i2+a2a1+a2b1i+b2a1i+b1b2i2=0 ⇒2a1a2−2b1b2+(2a1b2+2b1a2)i=0 ⇒2a1a2−2b1b2=0 and (2a1b2+2b1a2)=0 ⇒a1a2=b1b2