Is |a−iba+ib| equal to 1? If yes, how?

Answered

2022-01-17

Is |aiba+ib| equal to 1? If yes, how?

Answer & Explanation

nick1337

nick1337

Expert

2022-01-18Added 573 answers

Step 1
|z|=|z¯|
since the vector z¯ is a mirror copy of z , upon reflection into the real axis, which leaves its norm (length) unaltered.
The above assumes that a,bR. In fact the same is not true is we have a,bC
We do have
|z|2=zz¯
and
z1+z2¯=z1¯+z2¯z1z2¯=z1z2¯
But if we take a=1, b=i we find:
|1i×i|=|2|=2
and
|1+i×i|=|0|=0

Vasquez

Vasquez

Expert

2022-01-18Added 457 answers

Step 1 |aiba+bi| =|aib||a+ib| =a2+(b)2a2+b2 =1
alenahelenash

alenahelenash

Expert

2022-01-24Added 366 answers

Step 1Yes, presuming a and b are real. Then this is the quotient of conjugates, which have the same magnitude, so the magnitude of the quotient is 1. The result is on the unit circle, at double the angle of the numerator.|abia+bi|=|abi||a+bi|=a2+b2a2+b2=1If a and b are non-real complex numbers then in general this magnitude won't be 1.

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