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Raerin
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- Oct 7, 2013
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How do I solve z^4+16=0 in a+bi form. I don't see how it can be factored so what do I do?
Write it as:How do I solve z^4+16=0 in a+bi form. I don't see how it can be factored so what do I do?
I don't understand this part.Now you can consider \(\displaystyle 2(1+i)^2= 2(1+2i+i^2)=4i\)
I think my way is easier...I don't understand this part.
Is Serena's way easier? Though, I don't really know how to convert it into polar form.
(-16)^2 = sqrt(256) = 16?I think my way is easier...
Let's start with the modulus.
What is the modulus of $-16$?
Now suppose that the modulus of $z$ is $r$.
What should $r$ be?
Apply the expansion for $(x+y)^2=x^2+2xy+y^2$. Since we are looking at the roots of the equation \(\displaystyle z^4=-16\) and since $(z^2-4i)(z^2+4i)=0$ we have eitherI don't understand this part.
Is Serena's way easier? Though, I don't really know how to convert it into polar form.
Yep.(-16)^2 = sqrt(256) = 16?
r=2Yep.
Now we're looking for a modulus $r$ for $z$ such that $r^4 = 16$...
Exactly!
180 degrees?Exactly!
Now let's look at the angle.
Which angle would the number $-16$ have?
Right!180 degrees?
720?Right!
Now suppose $z$ has an angle $\phi$.
Then $z^4$ has an angle $4\phi$.
What should $\phi$ be?
Sorry, I still don't fully understand how z=2√(1+i)Apply the expansion for $(x+y)^2=x^2+2xy+y^2$. Since we are looking at the roots of the equation \(\displaystyle z^4=-16\) and since $(z^2-4i)(z^2+4i)=0$ we have either
\(\displaystyle z^2=4i\) or \(\displaystyle z^2=-4i\)
Now since for $z=\sqrt{2}(1+i)$ we have \(\displaystyle z^2=2(1+2i-1)=4i\) then
\(\displaystyle z=\sqrt{2}(1+i)\) is a solution. Similarily we have $z=-\sqrt{2}(1+i)$ as a solution. Using the same manner solve for $z$ the equation \(\displaystyle z^2=-4i\).
Not quite.720?
oh so the angle is 45?Not quite.
We have that $4\phi=180^\circ$.
Does that work with $\phi=720^\circ$?
Yes.oh so the angle is 45?
z = sqrt(2) + sqrt(2)i?Yes.
There you go. We now have 1 solution in polar form.
It has modulus $2$ and angle $45^\circ$.
Can you think of a complex number in the form $a+bi$ that corresponds to it?
Yes!z = sqrt(2) + sqrt(2)i?
Yep, I did! I just didn't understand how they got to that conclusion from z^2=4iYes!
Do you recognize the value for $z$ that ZaidAlyafey came up with?
Perhaps you might calculate $(\sqrt 2 + (\sqrt 2)i)^4$ to verify that it really is a solution?
Btw, there are 3 more solutions.
That is because $4\phi=180^\circ$ has more solutions.
If $4\phi = 360^\circ + 180^\circ$, we have the same angle but a different $\phi$...
Ah, I suspect he secretly converted to polar coordinates and back, and then presented the result.Yep, I did! I just didn't understand how they got to that conclusion from z^2=4i
Sort of...Squaring \(\displaystyle \sqrt{\frac{x}{2}}(1+i)\), we can generate any purely imaginary complex number.