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The square of what two complex numbers is 11+60i?

 
 Aug 6, 2022
 #1
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Hi Guest!

Let \(z=\sqrt{11+60i}\) where z is a complex number; hence of the form: \(z=x+iy\) where x and y are real numbers.

Thus, squaring both sides:

\((x+iy)^2=11+60i \\ \iff x^2-y^2+2xyi=11+60i \\ \text{Equating the real parts and imaginary parts, gives a system of two equations:} \\ x^2-y^2=11 \space \space \space \space \space \space (1)\\ 2xy=60 \implies xy=30 \space \space \space \space \space \space(2) \text{From (2), we make x or y the subject, let us make y the subject, then we substitute in (1)} \\ y=\dfrac{30}{x}. \space\\ \text{So, by (1):} \\ x^2-(\dfrac{30}{x})^2=11 \iff x^2-\frac{900}{x^2}=11 \iff x^4-900=11x^2 \iff x^4-11x^2-900=0\)

Solving for x (Use the substitution \(t=x^2\), then apply the quadratic formula) we get:

\(x_{1,2}=\pm6\)

So, \(y_{1,2}=\pm5\)   (By (2)).

Therefore, the two square roots are:

\(z_1=6+5i \\ z_2=-6-5i\)

I hope this helps!

 
 Aug 6, 2022
 #3
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Thank you!

 
Guest Aug 6, 2022
 #2
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Simplify the following:
sqrt(60 i + 11)

11 + 60 i = 36 + 60 i - 25 = 36 + 60 i + 25 i^2 = (5 i + 6)^2:
sqrt(((5 i + 6)^2))

Cancel exponents. sqrt((6 + 5 i)^2) = 5 i + 6:

6 + 5 i

 
 Aug 6, 2022

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