I was trying to prove thatwhere w is a complex number with absolute value smaller than 1 and r∈R, r&lt;1. I saw some solution online, which says we substitute w=meix so that the LHS becomes (m−r)2. How is that true?We have(w−r)(w―−r)=(meix−r)(me−ix−r)=eixe−ix(m−re−ix)(m−reix)=(m−re−ix)(m−reix)

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I was trying to prove thatwhere w is a complex number with absolute value smaller than 1 and r∈R, r&amp;lt;1. I saw some solution online, which says we substitute w=meix so that the LHS becomes (m−r)2. How is that true?We have(w−r)(w―−r)=(meix−r)(me−ix−r)=eixe−ix(m−re−ix)(m−reix)=(m−re−ix)(m−reix)

ze2m1ingkdvu · Accepted Answer

The assertion in the online solution appears to be incorrect.LHS=&amp;nbsp;|w|2−(w+w―)r+r2RHS=&amp;nbsp;1−(w+w―)r+|w|2r2Now compare&amp;nbsp;|w|2+r2&amp;nbsp;with&amp;nbsp;1+|w|2r2Evidence of the initial claim.1&amp;gt;r21−|w|2&amp;gt;r2(1−|w|2)=r2−r2|w|21+r2|w|2&amp;gt;r2+|w|2

Emelia Leon · Accepted Answer

LHS is not&amp;nbsp;(m−r)2, a trivial example would be w=i and r=1! In that case,&amp;nbsp;(m−r)2=0&amp;nbsp;which is apparently not true.Nevertheless, you can state that LHS:(ω−r)(ω―−r)=|ω−r|2=(mcos⁡(x)−r)2+(msin⁡(x))2

I was trying to prove that \(\) where w

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