Solving sqrt(-8.3)sqrt(1 - i8): Tips and Tricks for Complex Numbers"

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To solve sqrt(-8.3)sqrt(1 - i8), it's essential to recognize that you cannot separate the square roots of sums or differences. The correct approach involves simplifying the expression to a single square root, leading to z = (a + bi)^2, where you can equate real and imaginary parts. A common mistake is rewriting negative numbers as i^2, which can complicate the calculation. After simplifying, ensure to check for any sign errors in your calculations. Ultimately, solving the resulting equations will yield the correct complex number solution.
naspek
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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
 
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naspek said:
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,
\sqrt{a + b} \ne \sqrt{a} + \sqrt{b}
 
eumyang said:
Here's a problem (in bold). You can't take the square root of a sum/difference separately. In other words,
\sqrt{a + b} \ne \sqrt{a} + \sqrt{b}

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

First simplify the expression so that there is one square root. You sort of did that here (in bold):
naspek said:
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:
z = (a + bi)^2
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.
 
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