# Simple Complex number

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

eumyang
Homework Helper
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,
$\sqrt{a + b} \ne \sqrt{a} + \sqrt{b}$

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...

eumyang
Homework Helper
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):
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.

Last edited: