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complex Number
- Thread starter shaimaa
- Start date
find complex value in (a+bi) :
(z+2)^(1/2)/(3-2i)
what progress have you made so far?
do you know how to get a complex number out of the denominator?
do you know how to find the square root of a complex number?
yes, I multiply numerator and d enumerator by conjugate (3+2i)..
solve roots by polar form but , x&y make the problem very complex!!!
sounds to me like you know how to do it.
do you know how to back and forth between polar and cartesian forms?
sounds to me like you know how to do it.
do you know how to back and forth between polar and cartesian forms?
No, can you tell me?
No, can you tell me?
if you have z = (x + i y) then
z = r e^(i theta)
r = sqrt(x^2 + y^2)
theta = arctan(y/x)
you need to look at the x and y to get theta in the right quadrant, for example
1 + i = sqrt(2) e^(i pi/4)
whereas
-1 - i = sqrt(2) e^(i 5pi/4)
you need to understand this completely as complex analysis is full of converting back and forth between the two representations
Last edited:
if you have z = (x + i y) then
z = r e^(i theta)
r = sqrt(x^2 + y^2)
theta = arctan(y/x)
you need to look at the x and y to get theta in the right quadrant, for example
1 + i = sqrt(2) e^(i pi/4)
whereas
-1 - i = sqrt(2) e^(i 5pi/4)
you need to understand this completely as complex analysis is full of converting back and forth between the two representations
Ok , I understand all of this, but the presence of x &y ,make the problem very complex and difficult to solve.
Ok , I understand all of this, but the presence of x &y ,make the problem very complex and difficult to solve.
why?
z = x + i y
z + 2 = (x+2) + i y
now r = sqrt((x+2)^2 + y^2), theta = arctan(y/(x+2))
sqrt(z+2) = sqrt(r) e^(i theta/2)
etc etc
I suppose it's a bit of a mess but nothing you can't figure out
Last edited:
why?
z = x + i y
z + 2 = (x+2) + i y
now r = ((x+2)^2 + y^2), theta = arctan(y/(x+2))
sqrt(z+2) = sqrt(r) e^(i theta/2)
etc etc
I suppose it's a bit of a mess but nothing you can't figure out
sorry, (theta/2), how come?? and there is x & y..!!
why?
z = x + i y
z + 2 = (x+2) + i y
now r = ((x+2)^2 + y^2), theta = arctan(y/(x+2))
sqrt(z+2) = sqrt(r) e^(i theta/2)
etc etc
I suppose it's a bit of a mess but nothing you can't figure out
and r=sqrt((x+2)^2 +y^2)
and r=sqrt((x+2)^2 +y^2)
This is just how square roots work with complex numbers. You should know this.
if z = r e^(i theta) then sqrt(z) = sqrt(r) e^(i theta/2)
I guess you can show this as follows
let z = a b
then sqrt(z) = sqrt(a) sqrt(b)
z = r exp^(i theta)
so a = r, b = e^(i theta)
sqrt(a) = sqrt(r)
sqrt(b) = sqrt(e^(i theta)) = (e^(i theta))^1/2 = e^1/2(i theta) = e^(i theta/2)
so sqrt(z) = sqrt(r) e^(i theta/2)
>sorry , I thought you substituted by X&Y.. Thanks.This is just how square roots work with complex numbers. You should know this. if z = r e^(i theta) then sqrt(z) = sqrt(r) e^(i theta/2) I guess you can show this as follows let z = a b then sqrt(z) = sqrt(a) sqrt(b) z = r exp^(i theta) so a = r, b = e^(i theta) sqrt(a) = sqrt(r) sqrt(b) = sqrt(e^(i theta)) = (e^(i theta))^1/2 = e^1/2(i theta) = e^(i theta/2) so sqrt(z) = sqrt(r) e^(i theta/2)
D
Deleted member 4993
Guest
.why?
z = x + i y
z + 2 = (x+2) + i y
now r = ((x+2)^2 + y^2), theta = arctan(y/(x+2)) .............. shouldn't that be r^2 = (x+2)^2 + y^2
sqrt(z+2) = sqrt(r) e^(i theta/2)
etc etc
I suppose it's a bit of a mess but nothing you can't figure out
yep, my mistake, fixed.