21iz2-1=0
→z2=1/i
→z2=-i
→z2=1/2*(1-1-2i)
→z2=1/2*(1+i2-2i)
→z=1/√2*(1-i)
→|z|=1
if iz3 - z2 - z + i = 0, then show that |z| = 1
The solution given in my book is something like this...
(z - i)(iz2 - 1) = 0 so, z = i => |z| = 1 (i got it till here) but,
(iz2 - 1) = 0 then how can |z| = 1 ?? (plese help me understand this)!!!!
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6 Answers
21
262swaraj, you could have taken modulus right at the second step.
1isn`t the factoriszation wrong?
(z-i)(iz2-1) = iz3 - z +z2 + i but in the qstn it is -z2
36@Aditya -
maybe you mean this...
if iz2 - 1 = 0 then,
z2 = 1/i
=> |z|2 = 1/|i| => |z|2 = 1 or, |z| = 1 ?????
1iz2 - 1= 0 then,
z2= 1/i,
|z|2=1/|i|,
|z|2=1 implying |z|=1...