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Question: For all complex numbers z of the form\[1 + i\alpha \], if\[{z^2}{\text{ }} = {\text{ }}x{\text{ }} +...

For all complex numbers z of the form1+iα1 + i\alpha , ifz2 = x + iy{z^2}{\text{ }} = {\text{ }}x{\text{ }} + {\text{ }}iy, then:
A. y24x+2=0{y^2} - 4x + 2 = 0
B. y2+4x4=0{y^2} + 4x - 4 = 0
C. y24x+4=0{y^2} - 4x + 4 = 0
D. y2+4x+2=0{y^2} + 4x + 2 = 0

Explanation

Solution

It is given that z=1+iαz = {\text{1}} + i\alpha . Squaring it and equating it with  x + iy{\text{ }}x{\text{ }} + {\text{ }}iyis to be done by comparing real and imaginary parts.

Complete step-by-step answer:
We are given that a complex number is in the form z=1+iαz = {\text{1}} + i\alpha . It is also given that the square of this complex number is  x + iy{\text{ }}x{\text{ }} + {\text{ }}iy. It is shown as z2 = x + iy{z^2}{\text{ }} = {\text{ }}x{\text{ }} + {\text{ }}iy…… (1)
To solve the question first we need to find the square of the complex number.
z=1+iαz = {\text{1}} + i\alpha …… (2)
Squaring equation (2) both the sides we have,
z2=(1+iα)2=(1+iα)(1+iα){z^2} = {(1 + i\alpha )^2} = ({\text{1}} + i\alpha )({\text{1}} + i\alpha )
Solving it further we have,
z2=1+iα+iα+(iα)2{z^2} = 1 + i\alpha + i\alpha + {(i\alpha )^2}
z2=1+2iαα2{z^2} = 1 + 2i\alpha - {\alpha ^2} (i2=1{i^2} = - 1)
Rearranging the terms we get,
z2=(1α2)+(2α)i{z^2} = (1 - {\alpha ^2}) + (2\alpha )i……. (3)
Here 1α21 - {\alpha ^2} is the real part and 2α2\alpha is the imaginary part.
From equation (1) we have,
z2 = x + iy{z^2}{\text{ }} = {\text{ }}x{\text{ }} + {\text{ }}iy…….. (1)
Now equating both equation (1) and equation (3) we get,
x + yi=(1α2)+(2α)ix{\text{ }} + {\text{ yi}} = (1 - {\alpha ^2}) + (2\alpha )i
We have to equate the real part and imaginary part separately. Now equating we get,
xx= 1α21 - {\alpha ^2} ……. (4) And
y{\text{y}}=2α2\alpha ….. (5)
Here, the term alpha (α\alpha ) is parameter. A parameter is a changing variable. To get a relation between x and y as required we have to eliminate alpha (α\alpha ) from them.
Now from equation (5) we get,
y{\text{y}}=2α2\alpha
α\alpha = y2\dfrac{{\text{y}}}{2}….. (6)
Substituting value from equation (6) in equation (4) we have,
x\Rightarrow x= 1α21 - {\alpha ^2}
x\Rightarrow x= 1(y2)21 - {(\dfrac{y}{2})^2}
x=1y24\Rightarrow x = 1 - \dfrac{{{y^2}}}{4}
Multiplying both sides with and rearranging we get,
y2+4x4=0{y^2} + 4x - 4 = 0
Therefore, option (c) y2+4x4=0{y^2} + 4x - 4 = 0 is correct.

Note: When equating complex numbers we only need to equate the real part with the real and imaginary part with the imaginary part of the complex number. Complex numbers are the numbers that can be represented in the form of a+ib.