Solve for complex z equation z^2=2i?
- print Print
- list Cite
Expert Answers
calendarEducator since 2011
write5,348 answers
starTop subjects are Math, Science, and Business
You need to consider the following complex number `z = x + i*y` and you need to replace it in the given equation, such that:
`(x + i*y)^2 = 2i => x^2 + 2i*x*y + (i^2*y^2) = 2i`
Replacing `-1` for ` i^2` yields:
`x^2 - y^2 + 2i*x*y = 2i`
You need to equate the real parts both...
(The entire section contains 155 words.)
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
- How is solve in complex equation z+1/z=1?
- 1 Educator Answer
- Solve complex equationcomplex equation z^4+9=0 solve in z
- 1 Educator Answer
- Determine the modulus of the complex number z=(3-i)/(3-2i)
- 1 Educator Answer
- Solve the equation z + i = 2(1-z') z is a complex number
- 1 Educator Answer
- Solve the equation: |3x+2|=|3-2x|Solve the equation: |3x+2|=|3-2x|
- 1 Educator Answer
`z^2=2i`
Let
Z=a+ib
Thus
`(a+ib)^2=2i`
`a^2-b^2+2iab=2i`
comparing real and imaginary parts ,we have
`a^2-b^2=0`
`2ab=2`
`ab=1`
But
`(a^2+b^2)^2=(a^2-b^2)^2+4(ab)^2`
`(a^2+b^2)^2=4`
`a^2+b^2=sqrt(4)=2`
since a and b are real numbers . Also `a^2 ,b^2>=0` therefore
`a^2+b^2>=0` .It ruled out the possibilitt of negative root.
Thus
`a^2-b^2=0`
`a^2+b^2=2`
`=>2a^2=2`
`a=+-1`
`and b=+-1`
`Thus`
`z=-1-i`
`and `
`z=1+i`
Thus solutions of the equation are
(1+i) and (-1-i).
Student Answers