What does $\sqrt{3 + 7 i} \cdot {(12 + 5 i)}^{2}$ $sqrt(3+7i)*(12+5i)^2$ equal in a+bi form?

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What does $\sqrt{3 + 7 i} \cdot {(12 + 5 i)}^{2}$ $sqrt(3+7i)*(12+5i)^2$ equal in a+bi form?

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Answer 1 · 2017-08-28T07:27:39

$\sqrt{3 + 7 i} {(12 + 5 i)}^{2} = a + b i$ $sqrt(3+7i)(12+5i)^2=a+bi$ , where $a = 119 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} - 120 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$ $a=119sqrt(sqrt58/2+3/2)-120sqrt(sqrt58/2-3/2)$ and $b = 120 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + 119 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$ $b=120sqrt(sqrt58/2+3/2)+119sqrt(sqrt58/2-3/2)$

Explanation:

Let $\sqrt{3 + 7 i} = x + i y$ $sqrt(3+7i)=x+iy$ , then $3 + 7 i = (x^{2} - y^{2}) + 2 i x y$ $3+7i=(x^2-y^2)+2ixy$

Therefore $x^{2} - y^{2} = 3$ $x^2-y^2=3$ and $2 x y = 7$ $2xy=7$ and

${(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2} = 3^{2} + 7^{2} = 58$ $(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=3^2+7^2=58$

and $x^{2} + y^{2} = \sqrt{58}$ $x^2+y^2=sqrt58$

i.e. $2 x^{2} = \sqrt{58} + 3$ $2x^2=sqrt58+3$ and $x = \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}}$ $x=sqrt(sqrt58/2+3/2)$

and similarly $y = \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$ $y=sqrt(sqrt58/2-3/2)$

and $\sqrt{3 + 7 i} = \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + i \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$ $sqrt(3+7i)=sqrt(sqrt58/2+3/2)+isqrt(sqrt58/2-3/2)$

As ${(12 + 5 i)}^{2} = 144 - 25 + 120 i = 119 + 120 i$ $(12+5i)^2=144-25+120i=119+120i$

$\sqrt{3 + 7 i} {(12 + 5 i)}^{2} = (119 + 120 i) ⎛ ⎝ \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + i \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}} ⎞ ⎠$ $sqrt(3+7i)(12+5i)^2=(119+120i)(sqrt(sqrt58/2+3/2)+isqrt(sqrt58/2-3/2))$

= $119 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + 120 i \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + i 119 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}} - 120 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$ $119sqrt(sqrt58/2+3/2)+120isqrt(sqrt58/2+3/2)+i119sqrt(sqrt58/2-3/2)-120sqrt(sqrt58/2-3/2)$

= $⎡ ⎣ 119 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} - 120 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}} ⎤ ⎦ + i ⎡ ⎣ 120 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + 119 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}} ⎤ ⎦$ $[119sqrt(sqrt58/2+3/2)-120sqrt(sqrt58/2-3/2)]+i[120sqrt(sqrt58/2+3/2)+119sqrt(sqrt58/2-3/2)]$

Hence $\sqrt{3 + 7 i} {(12 + 5 i)}^{2} = a + b i$ $sqrt(3+7i)(12+5i)^2=a+bi$ , where $a = 119 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} - 120 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$ $a=119sqrt(sqrt58/2+3/2)-120sqrt(sqrt58/2-3/2)$ and $b = 120 \sqrt{\frac{\sqrt{58}}{2} + \frac{3}{2}} + 119 \sqrt{\frac{\sqrt{58}}{2} - \frac{3}{2}}$ $b=120sqrt(sqrt58/2+3/2)+119sqrt(sqrt58/2-3/2)$

Answer 2 · 2017-08-28T07:43:40

$\sqrt{3 + 7 i} {(12 + 5 i)}^{2}$ $sqrt(3+7i)(12+5i)^2$

$= (60 \sqrt{2 \sqrt{58} - 6} + \frac{119}{2} \sqrt{2 \sqrt{58} + 6}) + (60 \sqrt{2 \sqrt{58} + 6} - \frac{119}{2} \sqrt{2 \sqrt{58} - 6}) i$ $= (60 sqrt(2sqrt(58)-6) + 119/2 sqrt(2sqrt(58)+6))+(60 sqrt(2sqrt(58)+6) - 119/2 sqrt(2sqrt(58)-6))i$

Explanation:

Note that:

$| 3 + 7 i | = \sqrt{3^{2} + 7^{2}} = \sqrt{9 + 49} = \sqrt{58}$ $abs(3+7i) = sqrt(3^2+7^2) = sqrt(9+49) = sqrt(58)$

Since this is not a whole number, the square root of $3 + 7 i$ $3+7i$ has irrational real and imaginary parts. So it's not worth trying to guess.

Note that:

$3 + 7 i$ $3+7i$ is in Q1, so it has principal square root in Q1 and non-principal square root in Q3.
Its complex conjugate $3 - 7 i$ $3-7i$ is in Q4 with square roots in Q4 and Q2.

Note that if:

$x = \sqrt{3 + 7 i}$ $x = sqrt(3+7i)$

Then:

$x^{2} = 3 + 7 i$ $x^2 = 3+7i$

and:

${(x^{2} - 3)}^{2} = {(7 i)}^{2} = - 49$ $(x^2-3)^2 = (7i)^2 = -49$

So $\sqrt{3 + 7 i}$ $sqrt(3+7i)$ is the root in Q1 of:

$0 = {(x^{2} - 3)}^{2} + 49$ $0 = (x^2-3)^2+49$

$0 = x^{4} - 6 x^{2} + 9 + 49$ $color(white)(0) = x^4-6x^2+9+49$

$0 = x^{4} - 6 x^{2} + 58$ $color(white)(0) = x^4-6x^2+58$

Note that:

$(x^{2} - k x + \sqrt{58}) (x^{2} + k x + \sqrt{58}) = x^{4} + (2 \sqrt{58} - k^{2}) x^{2} + 58$ $(x^2-kx+sqrt(58))(x^2+kx+sqrt(58)) = x^4+(2sqrt(58)-k^2)x^2+58$

Equating coefficients:

$- 6 = 2 \sqrt{58} - k^{2}$ $-6 = 2sqrt(58)-k^2$

Hence:

$k^{2} = 2 \sqrt{58} + 6$ $k^2 = 2sqrt(58)+6$

So:

$k = \pm \sqrt{2 \sqrt{58} + 6}$ $k = +-sqrt(2sqrt(58)+6)$

Thus $\sqrt{3 + 7 i}$ $sqrt(3+7i)$ is the Q1 zero of:

$(x^{2} - \sqrt{2 \sqrt{58} + 6} x + \sqrt{58}) (x^{2} + \sqrt{2 \sqrt{58} + 6} x + \sqrt{58})$ $(x^2-sqrt(2sqrt(58)+6)x+sqrt(58))(x^2+sqrt(2sqrt(58)+6)x+sqrt(58))$

In order for the real part to be positive, we need the first of these two quadratics to be $0$ $0$ .

So using the quadratic formula with $a = 1$ $a=1$ , $b = - \sqrt{2 \sqrt{58} + 6}$ $b=-sqrt(2sqrt(58)+6)$ and $c = \sqrt{58}$ $c=sqrt(58)$ , we find:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{\sqrt{2 \sqrt{58} + 6} \pm \sqrt{(2 \sqrt{58} + 6) - 4 \sqrt{58}}}{2}$ $color(white)(x) = (sqrt(2sqrt(58)+6)+-sqrt((2sqrt(58)+6)-4sqrt(58)))/2$

$x = \frac{\sqrt{2 \sqrt{58} + 6}}{2} \pm \frac{\sqrt{2 \sqrt{58} - 6}}{2} i$ $color(white)(x) = sqrt(2sqrt(58)+6)/2+-sqrt(2sqrt(58)-6)/2i$

In order for the imaginary part to have poositive coefficient, we need the $+$ $+$ sign here.

The other factor is a little simpler:

${(12 + 5 i)}^{2} = (12^{2} - 5^{2}) + 2 (12) (5) i$ $(12+5i)^2 = (12^2-5^2)+2(12)(5)i$

${(12 + 5 i)}^{2} = (144 - 25) + 120 i$ $color(white)((12+5i)^2) = (144-25)+120i$

${(12 + 5 i)}^{2} = 119 + 120 i$ $color(white)((12+5i)^2) = 119+120i$

So:

$\sqrt{3 + 7 i} {(12 + 5 i)}^{2}$ $sqrt(3+7i)(12+5i)^2$

$= (\frac{\sqrt{2 \sqrt{58} + 6}}{2} + \frac{\sqrt{2 \sqrt{58} - 6}}{2} i) (119 + 120 i)$ $= (sqrt(2sqrt(58)+6)/2+sqrt(2sqrt(58)-6)/2i)(119+120i)$

$= (60 \sqrt{2 \sqrt{58} - 6} + \frac{119}{2} \sqrt{2 \sqrt{58} + 6}) + (60 \sqrt{2 \sqrt{58} + 6} - \frac{119}{2} \sqrt{2 \sqrt{58} - 6}) i$ $= (60 sqrt(2sqrt(58)-6) + 119/2 sqrt(2sqrt(58)+6))+(60 sqrt(2sqrt(58)+6) - 119/2 sqrt(2sqrt(58)-6))i$

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