Question

Solve this equation of complex numbers:

Answer 1

\[(z +2i)^{4}+4=0\]

Answer 2

You would need to expand the (z + 2i) 4 times. That's (z + 2i)(z + 2i)(z + 2i)(z + 2i). Do you know how to work with imaginary numbers? (z + 2i)(z + 2i) = z^2 + 4iz - 4. You need now to multiply that by itself so you FOIL it out the correct number of times. The result, before simplifying, is \[z ^{4}+4iz ^{3}-4z ^{?}+4iz ^{3}+16i ^{2}z ^{2}-16iz-4z ^{2}-16iz+16\]. This simplifies to \[z ^{4}+8iz ^{3}-32iz-8z ^{2}+16i ^{2}z ^{2}+16\]. And of course thei^2 terms can be further simplified to \[z ^{4}+8iz ^{3}-32iz-8z ^{2}-16z ^{2}+16\] One last simplification now. Combine the z^2 terms to get a final answer of \[z ^{4}+8iz ^{3}-32iz-24z ^{2}+16\]
WHEW!!!!!

Answer 3

But how do I find the solution of the equation? Isn't there a faster method?

Answer 4

Thank you for answering

Answer 5

Actually, no there isn't a simpler way. The whole idea of this type of problem is to be able to do this method with error. Unless you have a calculator that can solve the whole entire thing by plugging it in, this is it!

Answer 6

Ok, thanks. Regarding the solutions can you show me step by step how to do it? I'll give best answer for sure!

Answer 7

you may try :

\(\large z+2i = \left(4e^{i(\pi + 2k\pi)}\right)^{\frac{1}{4}} \)

\(k = 0,1,2,3\)

Answer 8

@ganeshie8 is this a particular formula that can be applied in similar cases?

Answer 9

thats just a method for finding \(n\)th roots

Answer 10

ah ok

Answer 11

thanks

Answer 12

\[(z +2i)^{4}+4=0\]
\[(z +2i)^{4}=-4\]

\[z+2i=\pm\sqrt[4]{-4}=\pm\sqrt{\pm2i}\]

\[\sqrt{2i}=\sqrt{(1+i)^2}=1+i\]
\[\sqrt{-2i}=i\sqrt{(1+i)^2}=i(1+i)=i-1\]
so

\[z=-2i\pm(1\pm i)\]

\[z=1-i,1-3i,-1-i,-1-3i\]

Answer 13

Can you explain to me the third and fourth passage? Where did the 1+i come from?

Answer 14

Thank you everybody for having helped me out.

Answer 15

yw