How $\sqrt{\cos (106.3) + i \sin (106.3)} = \cos 53.15 + i \sin 53.15$

Question

The question is find sqrt of $-7 +24i$ solution: $$\sqrt{-7+24i} = z$$ $$-7+24i = z^2$$ $r=25$, $106.3^\circ$

$$\sqrt{\cos (106.3) + i \sin (106.3)} = \cos 53.15 + i \sin 53.15 /*HOW?*/$$

thanks

Answer 1

De Moivre's formula $$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$$

$$\sqrt{\cos (106.3) + i \sin (106.3)} = \cos {1\over2}(106.3) + i \sin {1\over2}(106.3)$$

Answer 2

Using $e^{i \theta} = \cos \theta + i \sin \theta$ (Euler's formula), $$ \left(\cos \theta + i \sin \theta\right)^{1/2} = \left(e^{i \theta}\right)^{1/2} = e^{i \theta / 2} = \cos \left(\theta/2\right) + i \sin \left(\theta/2\right) $$

Answer 3

Let $z=x+iy$, and solve $z^2=-7+24i$: $$x^2-y^2=-7\\2xy=24,$$or, multiplying by $x^2$: $$x^4-x^2y^2=x^4-144=-7x^2.$$ This yields $$x^2=\frac{-7\pm\sqrt{625}}{2}=9,-16,$$then, keeping only the positive solution $x^2=9$, $$z=\pm(3+4i).$$