# How to solve the complex equation? $(x+2yi)^2 = xi.$

How to solve the following complex equation with in less than 60 seconds?

$$(x+2yi)^2 = xi.$$

I know how to solve, we have to solve power first then real part equal to real part and imaginary to imaginary.

Is there any simplest method than above to solve this one?

• I don't think the method you described should take very long. – Seth Jun 7 '14 at 11:42

I would follow exactly as you outlined using some simple tricks:

• The square of a complex number is complex
• $(a+b)^2 = a^2 + b^2 + 2ab$

Note that when you square $x + 2yi$ the only real terms will be $x^2$ and $(2yi)^2$ so add those and set them equal to zero off the bat and get a relation between $x,y$ $$x^2 - 4y^2 = 0 \implies x = \pm 2y$$ Then follow up with the complex equation $$4xyi = xi \implies y = 1/4 \implies x = \pm 1/2$$

Now I was able to do this while watching futurama and sleep deprived in about 100 seconds. I hope this provides at least some help!

• 100 seconds. In the exactly same way i have done this question. By the way thanks DanZimm. – zonnie Jun 7 '14 at 11:42
• @zonnie how long did it take you? – DanZimm Jun 7 '14 at 11:43
• You missed a solution. – Empy2 Jun 7 '14 at 11:44
• approximately same time DanZimm. – zonnie Jun 7 '14 at 11:45
• @Michael hrm let me guess, $x=y=0$? – DanZimm Jun 7 '14 at 11:46