# complex numbers?

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#1

1)simplify (express in form a+jb) cos3x+jsin3x/cosx +jsinx

2) if z=2+j/1-j find the real and imaginary parts of the complex number z+1/z

3)if x +y are real solve the equation

jx/1+jy = 3x +j4/x +3y

any advice would be much appreciated.
0
7 years ago
#2
For 1) write them in exponential form and then use normal exponential division laws.

For 2) Convert 1/z into a number with a real denominator and then add the real and imaginary parts separately to get a + jb

For 3) Convert the LHS to have a real denominator and equate coefficients

Quote me for any more help.
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#3
(Original post by tory88)
For 1) write them in exponential form and then use normal exponential division laws.

For 2) Convert 1/z into a number with a real denominator and then add the real and imaginary parts separately to get a + jb

For 3) Convert the LHS to have a real denominator and equate coefficients

Quote me for any more help.

any chance you could be more detailed? sorry i'm pretty new to complex numbers. for the first one do you mean using De moives formula.
0
7 years ago
#4
(Original post by Johnny185)
any chance you could be more detailed? sorry i'm pretty new to complex numbers. for the first one do you mean using De moives formula.
For the first one I mean rewrite it as r*e^(i*theta) where you can find r and theta from drawing out the diagram and using trigonometry/Pythagoras. If you show an attempt it would be easier for me to help.
0
7 years ago
#5
DeMoivre's formula does the same thing, since it is derived from Euler's formula that states cis(x)=e^ix
So yes, he means use DeMoivre's formula.
0
7 years ago
#6
(Original post by Johnny185)

1)simplify (express in form a+jb) cos3x+jsin3x/cosx +jsinx

2) if z=2+j/1-j find the real and imaginary parts of the complex number z+1/z

3)if x +y are real solve the equation

jx/1+jy = 3x +j4/x +3y

any advice would be much appreciated.
for 1)
To express in form a+bj multiply the numerator and denominator by cosx-jsinx
Expand the numerator and use addition rules for cos(A-B) and sin(A-B)

for 2)
z=(2+j)(1+j)/2->1/2+3/2j
z'=1/2-3/2j
1/z=z'/(z*z') ->Re(1/z) and Im(1/z)
then add together the real and imagine part of z and 1/z

for 3)
the form of equation is
Multiply both side by (1+jy)(x+3y) and expand both side
You will get two equations one for the real and one for the imagine part
Solve simultaneously for x and y
0
6 years ago
#7
Where would we be without Stroud?

For the last question multiply by the denominator, simplify then expand -: (1+jy)(x+3y)
jx(x+3y)=(3x+j4)(1+jy)
jx^2+j3xy=3x+j3xy+j4+j^24y
jx^2+j3xy=3x+j3xy+j4-4y

Collect in form 0=.....
0=(3x-4y)+j(3xy+4-x^2-3xy)
0=REAL (3x-4y)
0=IMAGINARY(4-x^2)

Solve
If 0 = 4-x^2 then x=srt(-4) or (x+2)(x-2)
And if =2 or -2
y=3x/4 ...
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