please help
Remember that..
|a + bi| = √(a² + b²)
Then...
a = 5
b = -√3
..you get...
√(5² + (-√3)²)
= √(25 + 3)
= √28
= 2√7
I hope this helps!
I z I = [ 25 + 3 ]^(1/2) = 28^(1/2) = 2â7
|a+bi| = sqrt(a^2+b^2)
|5-iâ3| = sqrt(5^2 + (-isqrt(3))^2) = 2sqrt(7)
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Verified answer
Remember that..
|a + bi| = √(a² + b²)
Then...
a = 5
b = -√3
..you get...
√(5² + (-√3)²)
= √(25 + 3)
= √28
= 2√7
I hope this helps!
I z I = [ 25 + 3 ]^(1/2) = 28^(1/2) = 2â7
|a+bi| = sqrt(a^2+b^2)
|5-iâ3| = sqrt(5^2 + (-isqrt(3))^2) = 2sqrt(7)