Question

by writing z in the form z=a+bi, find all solutions z of the following equations:
1) z^2=2+i
2) z^2-(3+i)z+(2+2i)=0

Answer 1

simplifying gives\[a^2-b^2+2abi=2+i\]\[a^2-b^2-3a+b+2+(2ab-a-3b+2)i=0\]

Answer 2

for example from first one \[a^2-b^2=2\]\[2ab=1\]combine 2 equations\[a^2-\frac{1}{4a^2}=2\]let \[t=a^2\]and solve

Answer 3

I do not understand..... plz show me step by step.

Answer 4

a^2-b^2+2abi=2+i
a^2-b^2= 2 , 2abi=i => 2ab=1
I got this part,


where did you get a^2-(1/4a^2)=2?

Answer 5

for the first one you are solving \(z^2=2+i\) for \(z\) meaning you need both square root sof \(2+i\)
if you are familiar with writing complex numbers in polar form, this would not be that hard.
otherwise you have \((a+bi)^2=a^2-b^2+2abi\) and solve \[a^2-b^2=2, 2ab=1\]
if you got that far, then \(2ab=1\implies b=\frac{1}{2a}\) making the first equation
\[a^2-\frac{1}{4a^2}=2\]

Answer 6

so for the first question. z=2. is that an answer?

Answer 7

this kind of sucks because you are going to have to solve
\[x^4-2x^2-\frac{1}{4}=0\]

Answer 8

no 2 is not and answer because \(2^2=4\) and not \(2+i\)

Answer 9

z1= (2+sqrt5)/2 , z2= (2-sqrt5)/2

Answer 10

@satellite73
is that a right answer?