Quick question - complex numbers (1 Viewer)

[skillz]

Someone please help me with the following complex numbers questions.


Simply each of the following for real values x and y

(x+yi)(3+2i)=-16+11i


(x+iy)^2=-18i

x+iy=(2y+1)+(x-7i)

Thank you for any help and your time.

 

[pLuvia]

(x+yi)(3+2i)=-16+11i
LHS = 3x + 3yi + 2xi - 2y
= 3x-2y + i(3y+2x)

Equate Re and Im parts
3x-2y=-16
3y+2x=11

Simlutaneously solve

x=-2
y=5

(x+iy)²=-18i
x<sup>2-y2+2xyi
Equate Re and Im

x2-y2=0
2xy=-18
y=-9/x

Solve

x4-81=0
x=3
y=-3</sup>

 

[Trev]

3.
I think you mean:
x+iy=(2y+1)+(x-7)i
Equating real and unreal parts:
2y+1=x
x-7=y
2(x-7)+1=x
&there4;x=13; y=6.

 

[skillz]

I'm still having a problem with this question.
(x+iy)2=-18i
I got up to.
x<sup>2-y2+2xyi
Equate Re and Im
x= -9i/y

(-9i/y)^2-y^2=0
(81i^2/y^2)-y^2=0

can someone explain to me what i'm doing wrong?

thank you.</sup>

 

[Riviet]

I'm still having a problem with this question.
(x+iy)2=-18i
I got up to.
x<sup>2-y2+2xyi
Equate Re and Im
x= -9i/y

(-9i/y)^2-y^2=0
(81i^2/y^2)-y^2=0

can someone explain to me what i'm doing wrong?

thank you.</sup>

^{When you are equating the imaginary part in a+ib, b is the imaginary part, not ib! So instead you should get x=-9/y and substitute that into the other equation.}

 

[skillz]

I see...

I'm at,
x^2-81/x^2=0

Do I times the X^2 by X^2 to get rid of X^2 underneath -81?

thank you again.

 

[Trev]

Yes, make it a simpler quadratic.

 

[skillz]

Thanks guys for the help.