Equation involved complex number

[alijan kk]

Homework Statement

Value of x and y , when (x+yi)²= 5+4i

Homework Equations

The Attempt at a Solution

x²+2x(iy)-y²=5+4i
x²-y²=5 -------> (1)
2x(iy)=4i (imaginary part)
xy=2 --------> (2)

solving the two equations

x=2.388 and y=0.838
or x=-2.388 or y=-0.838

is this the right way to solve it ?
its a multiple choice question and the options are:
x=2,y=-1
x=-2,y=1
x=2,y=-i
x=2,y=2

 

[Charles Link]

It looks to me like you solved it correctly. It looks like an error in the initial problem, like a missing minus sign somewhere.

 

[Ray Vickson]

Homework Statement

Value of x and y , when (x+yi)²= 5+4i

Homework Equations

The Attempt at a Solution

x²+2x(iy)-y²=5+4i
x²-y²=5 -------> (1)
2x(iy)=4i (imaginary part)
xy=2 --------> (2)

solving the two equations

x=2.388 and y=0.838
or x=-2.388 or y=-0.838

is this the right way to solve it ?
its a multiple choice question and the options are:
x=2,y=-1
x=-2,y=1
x=2,y=-i
x=2,y=2

Another (easier) way to get a numerical solution is to use a polar representation of complex numbers. If $x+iy = r e^{i \theta}$ then $(x + iy)^2 = r^2 e^{2 i \theta} = 5 + 4i$. We have $r^2 =\sqrt{ 5^2 + 4^2} = \sqrt{41},$ so $r = \sqrt{\sqrt{41}}$. Also, $\tan (2 \theta) = 4/5,$ so $\theta = (1/2) \arctan(4/5).$ You can evaluate $\theta, \sin(\theta)$ abd $\cos(\theta)$ on a calculator, and so get $x = r \cos(\theta), y = r \sin(\theta)$.