simple complex number (1 Viewer)

[hendo144]

i've been away sick from school lately and falling behind =(
how would u do this question:

find all real number a and b such that (a+ib)^2=9 +40i

cheers in advance

 

[annabackwards]

(a+ib)^2=9 +40i
a^2 - b^2 + 2abi = 9 + 40i
Therefore a^2 - b^2 = 9 and 2ab = 40 (by equating real values and imaginary values)

So use a^2 - b^2 = 9 and ab = 20 to solve for a and b simultaneously

 

[hendo144]

ty haha. heres another....
how would u solve z^2 + 2z(1+2i)- (11+2i)=0

 

[annabackwards]

I'd use the quadratic equation to find z.

You'll probably end up with a sqrt so to remove the square root, let (a + bi) = sqrt of whatever, square both sides and solve it simulatenously like the 1st question. Then substitute it back into the original quadratic and find z.

I'm a bit too tired to show you full working out, but i'll do it tomorrow arvo if you are having problems and no one else has posted up a solution by then

 

[addikaye03]

ty haha. heres another....
how would u solve z^2 + 2z(1+2i)- (11+2i)=0

Spoiler

Discriminant=b^2-4ac= (2+4i)^2-4(-11-2i)

=(4+16i+16i^2)+44+8i

=32+24i

Now, (a+ib)^2=32+24i [this part here can be is the same process as your original Q]

a^2+2iab+i^2b^2=32+24i

(a^2-b^2)+i(2ab)=32+24i

Now we equate Re(z) and Im(z):

a^2-b^2=32 (1) and ab=12 [b=12/a] (2)

Therefore (2 into 1):

a^2-(12/a)^2=32

a^4-32a^2-144=0 (factors 4,-36)

a^4-36a^2+4a^2-144=0

a^2(a^2-36)+4(a^2-36)=0

Since a,b E R a=+-6, so sub into 2 we find b=+-2

a+ib=+-(6+2i)

So now:

z=[-(2+4i)+-(6+2i)]/2 [quadratic formula]

=[-2-4i+2+6i]/2 or [-4-10i]/2

= i & -(2+5i)

Enjoy COMPLETE SOLUTION

Anymore Q let is know

 

[cutemouse]

ty haha. heres another....
how would u solve z^2 + 2z(1+2i)- (11+2i)=0

and then proceed as usual.

I have a horrible feeling that I made an arithmetic error somewhere, I'm not feeling too well today heh.

 

[jet]

and then proceed as usual.

I have a horrible feeling that I made an arithmetic error somewhere, I'm not feeling too well today heh.

Should be -12 under the square root sign in the fourth line

 

[addikaye03]

There is no errors in my solution is there Jetblack?

 

[elmoateme]

(a+ib)^2=9 +40i
a^2 - b^2 + 2abi = 9 + 40i
Therefore a^2 - b^2 = 9 and 2ab = 40 (by equating real values and imaginary values)

So use a^2 - b^2 = 9 and ab = 20 to solve for a and b simultaneously

Theres a slightly quicker way to solve than using these simultatenous equations that our class got taught.

Since (a^2 + b^2)^2 = a^4 + b^4 + 2a^2b^2
= (a^2-b^2)^2 + 4a^2b^2
= (a^2 -b^2)^2 + (2ab)^2
Now since a^2 - b^2 = 9 {eqn 1} and 2ab = 40 {eqn2}
therefore a^2 + b^2 = 41 {eqn 3}

now eqn1 + eqn 3
2a^2 =50
a=5
sub into eqn 2
b = 4

Initially I hated doing it this way, like most of our class, but after awhile we realised it was a lot easier in many cases since finding a is rather simple, and you make less stupid mistakes.

 

[hendo144]

thx guys for the help!

 

[jet]

There is no errors in my solution is there Jetblack?

Yeah, there is actually.

When you factorise the quartic in 'a', you mix up the signs of the factors - they should be +4 and -36. I haven't resubstituted my answers back into the original equation, but I'm pretty sure that will yield the right answers.

 

[addikaye03]

Yeah, there is actually.

When you factorise the quartic in 'a', you mix up the signs of the factors - they should be +4 and -36. I haven't resubstituted my answers back into the original equation, but I'm pretty sure that will yield the right answers.

shit, yeah fixed, i usually right PSF, just done it straight off computer. thanks for that, it will just yeild the wrong solutions for a and b, but overall answer would be the same.