Need clarification on how to proceed with this.

Question

Can you help me in solving this equation? Thanks in advance :)

$(x + iy)^2 = 5 + 4i$

You are asked to get the square roots of $5+4i.$ See http://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number — mfl, Dec 08 '16 at 18:05
In the future: titles should be descriptive. You LaTeX markup. Give us info about what methods or attempts you know or have tried. You'll get more help that way. — The Count, Dec 08 '16 at 18:09

score 2 · Answer 1 · edited Dec 08 '16 at 18:19

2

$x^2 - y^2 + 2ixy = 5 + 4i$

Equating real and imaginary parts

$x^2-y^2 = 5$

$xy = 2$

That should be enough, right ?

edited Dec 08 '16 at 18:19

Harsh Kumar

2,846

answered Dec 08 '16 at 18:09

Vishnu V.S

975

Please use LaTeX markup in the future. – The Count Dec 08 '16 at 18:09
Note the OP's earlier question, http://math.stackexchange.com/questions/2049848/how-to-solve-this-equation-x2-y2-5-xy-2-find-x-and-y – Barry Cipra Dec 08 '16 at 18:26

score 1 · Answer 2 · answered Dec 08 '16 at 18:07

1

$$ x^2+2xyi-y^2=5+4i$$ $$x^2-y^2=5\tag1$$ $$2xy=4 \tag2$$ then solve the two equations

answered Dec 08 '16 at 18:07

E.H.E

23,280

score 1 · Answer 3 · edited Dec 12 '16 at 06:36

1

The square root is not a well defined function on complex numbers. If you want to find out the possible values, the easiest way is probably to go with "De Moivre's formula", that is, converting your number into the form

$$r(\cos(\theta) + i \sin(\theta))$$ and then you will get,

$$(r(\cos(\theta)+ i \sin(\theta)))^{1/2} = ±[\sqrt{r}(\cos(\theta/2) + i \sin(\theta/2))]$$.

edited Dec 12 '16 at 06:36

Vidyanshu Mishra

10,143

answered Dec 08 '16 at 18:12

Harsh Kumar

2,846

Need clarification on how to proceed with this.

3 Answers3