# Square root of complex numbers... I need... help... (1 Viewer)

#### zoeeoz18

##### New Member
SQUARE ROOT OF COMPLEX NUMBERS.... I need help...

ok so, I'm having a lot of trouble with this area, does anyone happen to have a good way of solving them? I've tried the quadratic formula one but I keep getting stuck at a certain point

Finding the square roots of 2-√5i is definitely one of the questions that I keep getting wrong.

#### psmao

##### Member
for these type of questions, you just need "find z strategy".
let $\bg_white z=x+iy, z^{2}=2-\sqrt(5)i$
and then expand, by equating the real parts and imaginary parts you should be able to get two equations about x and y
now all you need to do is to solve simultaneously.

try some exercise and hopefully you can figure it out.
I was confused initially as well but after I realised complex number is basically just let z=x+iy, then it becomes so easy to me.

this technique can also be used when sketching them but it will come later.

GLHF

#### Skuxxgolfer

##### Member
drop 4u then, complex numbers is one of the easiest topics
Its not the easiest topic. The questions range from extremely easy to impossibly difficult

#### Drongoski

##### Well-Known Member
I like the method in Terry Lee's text:

$\bg_white z^2 = 2 - \sqrt 5 i = \frac {1}{2}(4 - 2\sqrt 5 i) = \frac {1}{2}(5 - 2\sqrt 5 i -1) \\ \\ = \frac{1}{2}((\sqrt 5)^2 +2\times\sqrt 5 \times (-i) +(-i)^2)) = \frac{1}{2} (\sqrt 5 - i)^2\\ \\ \therefore z = \pm \frac {1}{\sqrt 2} (\sqrt 5 - i)$

@Zoee: You may find this solution more confusing than helpful at this stage of your learning complex numbers. But, when you finally get on top of it, you'll find this method elegant and brief. But you have to be good in algebraic fiddling.

Last edited:

#### Heresy

##### Active Member
- This should hopefully help

#### Drongoski

##### Well-Known Member
$\bg_white -5 + 12i = 4 + 2 \times 2 \times 3i -9 = 2^2 + 2\times 2 \times 3i + (3i)^2 = (2 + 3i)^2 \\ \\ \therefore \sqrt {-5 + 12i} = \pm(2 + 3i)$

Take your pick: which is shorter??

#### Arrowshaft

##### Well-Known Member
SQUARE ROOT OF COMPLEX NUMBERS.... I need help...

ok so, I'm having a lot of trouble with this area, does anyone happen to have a good way of solving them? I've tried the quadratic formula one but I keep getting stuck at a certain point

Finding the square roots of 2-√5i is definitely one of the questions that I keep getting wrong.

I'm gonna teach you the 'normal' way of doing it, Drongoski showed you the shortcut, I'm building off of Psmao's method.

So, we are given that

$\bg_white z^2=2-\sqrt{5}i$

Now remember that the complex number $\bg_white z$ can be expressed as $\bg_white z=x+iy$ where $\bg_white x,y\in\mathbb{R}$,

$\bg_white (x+iy)^2=2-\sqrt5i$

$\bg_white x^2+2ixy+i^2y=-2-\sqrt5i$

$\bg_white \left(x^2-y^2\right)+(2xy)i=2-\sqrt{5}i$

Now, equate the real and imaginary components, first the real ones:

$\bg_white x^2-y^2=2~~(1)$

Imaginary:

$\bg_white 2xy=-\sqrt{5}~~(2)$

From $\bg_white (2)$ we see that,

$\bg_white y=-\dfrac{\sqrt{5}}{2x}~~(3)$

Now substitute this into $\bg_white (1)$ to yield,

$\bg_white x^2-\dfrac{5}{4x^2}=2$

$\bg_white 4x^4-5=8x^2$

$\bg_white 4x^4-8x^2-5=0$

Note that the polynomial above is a quadratic in $\bg_white x^2$, so use the quadratic formula,

$\bg_white x^2=\dfrac{8\pm\sqrt{8^2-4\times4\times-5}}{8}=\dfrac{8\pm12}{2}$

Since $\bg_white x,y\in\mathbb{R}$, this means that we take the positve solution

$\bg_white x=\pm\sqrt{\dfrac{20}{8}}=\pm\sqrt{\dfrac{5}{2}}$

From $\bg_white (3)$ this means,

$\bg_white y=\mp\dfrac{\sqrt{5}}{2\times\dfrac{\sqrt5}{\sqrt2}}$

$\bg_white y=\mp\dfrac{\sqrt2}{2}=\mp\dfrac{1}{\sqrt2}$

Therefore,

$\bg_white z=\pm\left(\dfrac{\sqrt5}{\sqrt2}-\dfrac{1}{\sqrt2}i\right)$

$\bg_white z=\pm\dfrac{1}{\sqrt2}\left(\sqrt5-i\right)$

Last edited:

#### Drongoski

##### Well-Known Member
Well - you can compare the "normal" method with the "abnormal" method. If you can manage the shorter method, would you want to go through the rigmarole of the "normal" method, unless the question mandates this approach.

#### Arrowshaft

##### Well-Known Member
Well - you can compare the "normal" method with the "abnormal" method. If you can manage the shorter method, would you want to go through the rigmarole of the "normal" method, unless the question mandates this approach.
agreed. Just wanted to give him an idea of the other option

#### Drongoski

##### Well-Known Member
I didn't mean to criticise your method. It is, after all, the standard method. But for simple cases, the "Terry Lee" method is really short. But I bet many will find it hard, unless they are handy with algebraic fiddling.

Last edited: