Simple Complex number

  • Thread starter naspek
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  • #1
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how to solve sqrt(-8.3)sqrt(1 - i8)?


i try to solve it.. but got the wrong answer..

sqrt(-8.3)sqrt(1 - i8) = sqrt[(8.3i^2)(1 - 8i)]
= sqrt (8.3i^2 - 66.4i)
= 2.88i + 8.15

the answer should be.. 5.41 + i6.13
 

Answers and Replies

  • #2
eumyang
Homework Helper
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how to solve sqrt(-8.3)sqrt(1 - i8)?


i try to solve it.. but got the wrong answer..

sqrt(-8.3)sqrt(1 - i8) = sqrt[(8.3i^2)(1 - 8i)]
= sqrt (8.3i^2 - 66.4i)
= 2.88i + 8.15
Here's a problem (in bold). You can't take the square root of a sum/difference separately. In other words,
[itex]\sqrt{a + b} \ne \sqrt{a} + \sqrt{b}[/itex]
 
  • #3
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Here's a problem (in bold). You can't take the square root of a sum/difference separately. In other words,
[itex]\sqrt{a + b} \ne \sqrt{a} + \sqrt{b}[/itex]

then... what should i do..? i got stuck there...
 
  • #4
eumyang
Homework Helper
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Assuming you want just the principal square root, consider this: if there is a complex number a + bi such that
[itex]\sqrt{z} = a + bi[/itex],
then it makes sense that
[itex]z = (a + bi)^2[/itex].

First simplify the expression so that there is one square root. You sort of did that here (in bold):
sqrt(-8.3)sqrt(1 - i8) = sqrt[(8.3i^2)(1 - 8i)]
= sqrt (8.3i^2 - 66.4i)
= 2.88i + 8.15
... but there is a sign mistake. Also, forget about rewriting a negative as i2 in your 1st step.

Whatever is under the square root is your z. Take this:
[itex]z = (a + bi)^2[/itex]
and expand the right-hand side. Equate the real number parts and the imaginary number parts. You'll end up with 2 equations and 2 unknowns. Solve for a and b.
 
Last edited:

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