Where a and b are real numbers, It is still impossible to find the square root of i because it is imaginary. However, your solutions will be given as: z₀ = 0.06 + 5 * i; and z₁ ≈ 0.0562 + 1*i.
What is the explanation for the above solution?
z² = a + b x i
z² = -7+24i ; where a = -7; and b = 24
z = √(-7+ 24i)
r = √(a² + b²)
r = √(-7² + 24²)
r = √(49+576)
r=√629
r = 25
tan [tex]\varphi[/tex]= b/a
tan [tex]\varphi[/tex]= 24/-7
tan [tex]\varphi[/tex] = - 3.42857142857
[tex]\varphi[/tex] = arctan( - 3.42857142857) + 180°
[tex]\varphi[/tex]= -1.28700222 + 180°
[tex]\varphi[/tex] = 178.71299778
z = [tex]\sqrt{r} *\left[ cos (\frac{\varphi}{2} +k * \frac{360}{2}) + i*sin (\frac{\varphi}{2} + k * \frac{360}{2}) \right][/tex]
z₀ = [tex]\sqrt{25} *\left[ cos (\frac{178.71299778}{2} +0 * \frac{360}{2}) + i*sin (\frac{178.71299778}{2} + 0 * \frac{360}{2}) \right][/tex]
z₀ = 5 * [tex]\left[ cos (\fra{89.35649889)} + i*sin ({89.35649889} }) \right][/tex]
z₀ = 5 * cos 89.35649889 + i * 5 * sin (89.35649889)
z₀ = 0.056154884967 + 4.9996846529 * i
z₀ = 0.06 + 5 * i
z₁ = [tex]\sqrt{25} *\left[ cos (\frac{178.71299778}{2} + 1 * \frac{360}{2}) + i*sin (\frac{178.71299778}{2} + 1 * \frac{360}{2}) \right][/tex]
z₁ = (5 * 0.0112309769934135765421929926216) + i * 0.99993693058901140624626993438398
z₁ ≈ 0.0562 + 1*i
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