Find the complex square roots of 5-12i by solving (x+iy)^2=5-12i for x,y real. Thanks!
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Originally Posted by Rine198 Find the complex square roots of 5-12i by solving (x+iy)^2=5-12i for x,y real. $\displaystyle (x+iy)^2=x^2-y^2+2xyi=5-12i$ Now, identify real and imaginary parts and solve the system.
so i've identified that 5=x^2-y^2 (real part) -12=2xy (imaginary part) and i tried to do simultaneous equation with it, but i still don't get how i can get the answer?
Originally Posted by Rine198 5=x^2-y^2 (real part) -12=2xy (imaginary part) and i tried to do simultaneous equation with it, but i still don't get how i can get the answer? Right, now $\displaystyle y=-6/x$, substitute in the first equation and you'll obtain a biquadratic equation on $\displaystyle x$ .
Originally Posted by Rine198 so i've identified that 5=x^2-y^2 (real part) -12=2xy (imaginary part) and i tried to do simultaneous equation with it, but i still don't get how i can get the answer? Show what you have done.
but how do i factorise x^4-36-5x^2=0, which is what i got after substituting y=-6/x
Let $\displaystyle u = x^2$ such that $\displaystyle x^4-5x^2-36 = u^2-5u-36 = 0$ This one does factorise - what two number multiply to -36 and sum to -5?
So i've found that x=+/-3 and y=+/-2, but i still don't get how to find the complex square roots from this?
The square roots are $\displaystyle 3-2i,-3+2i$ . Equivalently : $\displaystyle \pm(3-2i)$ .
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