a+in=√1+i÷1-i,prove that a^2+b^2=1 ​

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a+in=√1+i÷1-i,prove that a^2+b^2=1
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Ben Gia 3 years 2021-08-02T06:32:51+00:00 1 Answers 3 views 0

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  1. thachthao
    0
    2021-08-02T06:34:35+00:00
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    Answer with Step-by-step explanation:

    We are given that

    a+ib=\sqrt{\frac{1+i}{1-i}}

    We have to prove that

    a^2+b^2=1

    a+ib=\sqrt{\frac{(1+i)(1+i)}{(1-i)(1+i)}}

    Using rationalization property

    a+ib=\sqrt{\frac{(1+i)^2}{(1^2-i^2)}}

    Using the property

    (a+b)(a-b)=a^2-b^2

    a+ib=\sqrt{\frac{(1+i)^2}{(1-(-1))}}

    Using

    i^2=-1

    a+ib=\frac{1+i}{\sqrt{2}}

    a+ib=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}

    By comparing we get

    a=\frac{1}{\sqrt{2}}, b=\frac{1}{\sqrt{2}}

    a^2+b^2=(\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{2}})^2

    a^2+b^2=\frac{1}{2}+\frac{1}{2}

    a^2+b^2=\frac{1+1}{2}

    a^2+b^2=\frac{2}{2}

    a^2+b^2=1

    Hence, proved.

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