could someone answer these simple questions plz? (1 Viewer)

[prsce24]

i just started learning complex no. and don't get these questions:

a) find reciprocal of i


answer = 1


b) Find and express z=a+ib such that Re(z)=2 and z^2 is imaginary

answer = z=2+2i & 2-2i


this is all from the cambridge book
thanx guys

 

[withoutaface]

recip of i=1/i=i/-1=-i

If Re(z)=2 then a=2

If Re(z^2)=0

:. a^2-b^2=0
:. a^2=b^2
:. 4=b^2
:.+/-2=b

:.z=2+2i or 2-2i

 

[McLake]

a) The reciprical of i is 1/i
1/i = 1/i * i/i = i/-1 = -i

typo?

b) a must be 2.
so z^2 = (2 + ib)^2 = 4 + 4ib - b^2
so 4 - b^2 = 0
b^2 = 4
b = +/- 2

so z = 2 + 2i or 2 - 2i

 

[prsce24]

dunno if its typo
it says

"i(-i)=1
i and -i are reciprocals

 

[prsce24]

so z^2 = (2 + ib)^2 = 4 + 4ib - b^2
so 4 - b^2 = 0

i get upto 4 - b^2 +4ib
but why do we only take 4 - b^2 = 0
why dont we take 4 - b^2 + 4ib=0?

 

[McLake]

dunno if its typo
it says

"i(-i)=1
i and -i are reciprocals

So -i is the reciprical, not 1.

so z^2 = (2 + ib)^2 = 4 + 4ib - b^2
so 4 - b^2 = 0

i get upto 4 - b^2 +4ib
but why do we only take 4 - b^2 = 0
why dont we take 4 - b^2 + 4ib=0?

you know that z^2 is Imginary. So the real part must be 0. we know that b is real (by definition), so the real part is 4 - b^2

 

[withoutaface]

The reason it says that is because, by definiton an number*its reciprocal=1

 

[prsce24]

oh ok i get it now : )
thanx u two.