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1/(2 + 5i) (multiply both the numerator and the denominator by 2 - 5i)
= 1(2 - 5i)/(2 + 5i)(2 - 5i)
= (2 - 5i)/(4 - 25i2) (substitute -1 for i2)
= (2 - 5i)/(4 + 25)
= (2 - 5i)/29
= 2/29 - (5/29)i

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Q: What is 1 divided by 2 plus 5i in the form a plus bi?
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What is the standard form of negative square root of 5i?

The standard form of a complex number is a+bi. So the standard form of the negative square root of 5i is 0-√(5i).


What is the reciprocal of 2-5i?

Let a + bi be the reciprocal. So (a + bi)(2 - 5i) = 1 (a + bi)(2 - 5i) = 2a - 5ai + 2bi + 5b = (2a + 5b) + (2b - 5a)i Therefore 2a + 5b = 1 and 2b - 5a = 0. Solving the simultaneous equations, we find that a = 2/29 and b = 5/29. So the reciprocal of 2 - 5i is 2/29 + 5i/29.


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