how can i solve this: z^2=9-40i
- Thread starter oded244
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well, to be honest im not from an english spechink country so i dont know all the names for the subjects where learning so its pretty hard for me to explain.
what i've done:
z^2=9-40i
z=sqrt 9-40i
sqrt 9-40i=a+bi ^2
9-40i=a^2+2abi-b^2
9=a^2-b^2
-40=2ab
-20/b = a - im stuck here, when i place the a i've got in the equation above i get a wrong answer
what i've done:
z^2=9-40i
z=sqrt 9-40i
sqrt 9-40i=a+bi ^2
9-40i=a^2+2abi-b^2
9=a^2-b^2
-40=2ab
-20/b = a - im stuck here, when i place the a i've got in the equation above i get a wrong answer
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- Feb 4, 2004
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I don't see how you got this line...?oded244 said:
I will guess that you're trying to find the value of the complex number "z". Since z = a + bi for some real values a and b, try working directly from this:
. . .z<sup>2</sup> = (a + bi)<sup>2</sup>
. . . . .= a<sup>2</sup> + 2abi + b<sup>2</sup> i<sup>2</sup>
. . . . .= a<sup>2</sup> + 2abi - b<sup>2</sup>
. . . . .= (a<sup>2</sup> - b<sup>2</sup>) + 2abi
...where a<sup>2</sup> - b<sup>2</sup> = 9 and 2ab = -40.
Solve this system for the values of a and b. :wink:
Eliz.
Hello, oded244!
Here's one method . . .
Let \(\displaystyle z \:=\:a\,+\,bi\;\) where \(\displaystyle a\) and \(\displaystyle b\) are real numbers.
Then we have: \(\displaystyle \
a\,+\,bi)^2 \:=\:9\,-\,40i\)
Expand and we get: \(\displaystyle \a^2\,-\,b^2)\,+\,2abi \:=\:9\,-\,40i\)
Equate real and imaginary components and we have:
. . \(\displaystyle \begin{array}{cccc}a^2\,-\,b^2 & \:=\: & 9 & \;[1]\\ 2ab & \:=\:&-40 & \;[2]\end{array}\)
From [2], we have: \(\displaystyle \:b \:=\:-\frac{20}{a}\)
Substitute into [1]: \(\displaystyle \:a^2\,-\,\left(-\frac{20}{a}\right)^2 \:=\:9\;\;\Rightarrow\;\;a^2\,-\,\frac{400}{a^2}\:=\:9\)
. . \(\displaystyle a^4\,-\,9a^2\,-\,400\:=\:0\;\;\Rightarrow\;\;(a^2\,-\,25)(a^2\,+\,16)\:=\:0\)
And we have: \(\displaystyle \:a \:=\:\pm5,\;\;b \:=\:\mp4\)
Therefore: \(\displaystyle \:\L\fbox{z \;=\;\pm(5\,-\,4i)}\)
Here's one method . . .
Let \(\displaystyle z \:=\:a\,+\,bi\;\) where \(\displaystyle a\) and \(\displaystyle b\) are real numbers.
Then we have: \(\displaystyle \
a\,+\,bi)^2 \:=\:9\,-\,40i\)Expand and we get: \(\displaystyle \
a^2\,-\,b^2)\,+\,2abi \:=\:9\,-\,40i\)Equate real and imaginary components and we have:
. . \(\displaystyle \begin{array}{cccc}a^2\,-\,b^2 & \:=\: & 9 & \;[1]\\ 2ab & \:=\:&-40 & \;[2]\end{array}\)
From [2], we have: \(\displaystyle \:b \:=\:-\frac{20}{a}\)
Substitute into [1]: \(\displaystyle \:a^2\,-\,\left(-\frac{20}{a}\right)^2 \:=\:9\;\;\Rightarrow\;\;a^2\,-\,\frac{400}{a^2}\:=\:9\)
. . \(\displaystyle a^4\,-\,9a^2\,-\,400\:=\:0\;\;\Rightarrow\;\;(a^2\,-\,25)(a^2\,+\,16)\:=\:0\)
And we have: \(\displaystyle \:a \:=\:\pm5,\;\;b \:=\:\mp4\)
Therefore: \(\displaystyle \:\L\fbox{z \;=\;\pm(5\,-\,4i)}\)
D
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oded244 said:
Another method would be to use DeMoivre's (sp) Theorem - where the complex numbers expressed as exponential functions.