Math, asked by Shashi565, 11 months ago

find z if arg(z+2i)=π/4 and arg(z-2i)=3π/4 solutions


brunoconti: are you sure about this problem? i suspect that something is wrong?

Answers

Answered by MaheswariS
2

\underline{\textsf{Given:}}

\mathsf{arg(z+2i)=\dfrac{\pi}{4},\;arg(z-2i)=\dfrac{3\pi}{4}}

\underline{\textsf{To find:}}

\textsf{z}

\underline{\textsf{Solution:}}

\textsf{Let z=x+iy}

\mathsf{arg(z+2i)=\dfrac{\pi}{4}}

\mathsf{arg(x+iy+2i)=\dfrac{\pi}{4}}

\mathsf{arg(x+i(y+2))=\dfrac{\pi}{4}}

\mathsf{tan^{-1}(\dfrac{y+2}{x})=\dfrac{\pi}{4}}

\mathsf{\dfrac{y+2}{x}=tan\dfrac{\pi}{4}}

\mathsf{\dfrac{y+2}{x}=1}

\mathsf{y+2=x}.....(1)

\textsf{and}

\mathsf{arg(z-2i)=\dfrac{3\pi}{4}}

\mathsf{arg(x+iy-2i)=\dfrac{3\pi}{4}}

\mathsf{arg(x+i(y-2))=\dfrac{3\pi}{4}}

\mathsf{tan^{-1}(\dfrac{y-2}{x})=\dfrac{3\pi}{4}}

\mathsf{\dfrac{y-2}{x}=tan\dfrac{3\pi}{4}}

\mathsf{\dfrac{y-2}{x}=-1}

\mathsf{y-2=-x}........(2)

\textsf{Adding (1) and (2), we get}

\mathsf{2y=0}

\implies\mathsf{y=0}

\textsf{Put y=0 in (1), we get}

\textsf{x=2}

\implies\boxed{\mathsf{z=2+i0}}

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