z^2=7-6i; solve for z

Question

wmj259 · Answer

assume z is y 
and i is x

wmj259 · Answer

$$y ^{2}=7-6x$$

wmj259 · Answer

$$\sqrt{y ^{2}}=\sqrt{7-6x}$$

wmj259 · Answer

$$y=\sqrt{7-6x}$$

wmj259 · Answer

Just switch y for z 
and 
x for i

thesmartone · Answer

So after doing what @wmj259 did you get
$$z=\sqrt{7-6i}$$

paxpolaris · Answer

we can do better than that, no?

thesmartone · Answer

I don't think so...

geerky42 · Answer

Why is it necessary to replace z and i to y and x respectively?

freckles · Answer

$$\text{ well you can write } 7-6i \text{ in } re^{i \theta} \text{ form }$$$$z^2=r (\cos(\theta)+i \sin(\theta))=r(\cos( \theta+ 2 \pi n)+i \sin(\theta+2 \pi n)) =re^{i (\theta+2 \pi n)} \ z=r^\frac{1}{2}e^{\frac{1}{2}i (\theta+ 2 \pi n)}$$
And we only need to two z values since we are finding square root
so n=0,1 in this case

freckles · Answer

All you have to do is find r and thet

paxpolaris · Answer

|dw:1414883861099:dw|
$$y^2=7-6i = \sqrt{85}e^{i \theta}$$


i guess for the exact form of y we'd have to leave the expression as$$\large y = \color{red}\pm \sqrt{7-6i}$$

wmj259 · Answer

@geerky42, you don't have the change the variables, but its easier to see it.

wmj259 · Answer

that is true that when you take the square root you have to include the plus or minus

paxpolaris · Answer

the approximate form would be:$$y \approx \pm2.8478\mp1.0535 i$$