What complex numbers have the same absolute value as #sqrt(3)+i# but subtend a right angle with it at O in the complex plane?

1 Answer
George C.
Oct 31, 2016

(2) #" "-1+isqrt(3)" "# or #" "1-isqrt(3)#

Explanation:

Assuming 'O' is #0#, i.e. the origin, we are basically asking what Complex numbers do you get from #sqrt(3)+i# by rotating by a right angle - clockwise or anticlockwise - about #0#.

Rotating anticlockwise about #0# by a right angle is the same as multiplying by #i = cos(pi/2)+i sin(pi/2)#:

#i(sqrt(3)+i) = isqrt(3)+i^2 = isqrt(3)-1 = -1+isqrt(3)#

Rotating clockwise about #0# by a right angle is the same as multiplying by #-i = cos(-pi/2)+i sin(-pi/2)#:

#-i(sqrt(3)+i) = -isqrt(3)-i^2 = -isqrt(3)+1 = 1-isqrt(3)#

So the answer is:

(2) #" "-1+isqrt(3)" "# or #" "1-isqrt(3)#

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