How do you simplify #abs(-12 + 22.5i)#?

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Answer 1 · 2016-01-10T10:18:56

Please go through the steps given below.

If #z=a+ib# is a complex number then #|z| = sqrt(a^2+b^2)#

Our problem is #|-12+22.5i|#

The modulus is found by the rule mentioned above

#|-12+22.5i| = sqrt((-12)^2+(22.5)^2)#

#|-12+22.5i| = sqrt(144+506.25)#

#|-12+22.5i| = sqrt(650.25)#

#|-12+22.5i| = 25.5#