Posted by **herp_derp** on Monday, March 10, 2014 at 11:12pm.

Please help me with this problem.

Simplify.

√(-24 - 10i)

- Algebra II -
**herp_derp**, Monday, March 10, 2014 at 11:17pm
oh, and also, if it is not too much too ask, please include steps and explanations. ty

- Algebra II -
**Steve**, Monday, March 10, 2014 at 11:48pm
√(-24 - 10i)

If you want to find a perfect square which equals -24-10i, it will have to include -5i, so

(a-5i)^2 = -24-10i

a^2 - 10ai - 25 = -24-10i

Hmmm. Looks like a=1

(1-5i)^2 = 1 - 10i - 25 = -24-10i

So,

√(-24 - 10i) = 1-5i or -1+5i

Or, if you use polar form, you can see that

r=26

tanθ = 5/12

the square root has

r = √26

tanθ = (1-(-12/13))/(-5/13) = -5

so, we get the same result.

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