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Convert of the complex number in the polar form:`1" "" "i`

Answer» The given complex number is z = -1 -i.
Let its polar form be `z = r(cos theta + i sin theta)`.
Now, `r=|z|=sqrt((-1)^(2)+(-1)^(2))=sqrt(2)`.
Let `alpha` be the acute angle, given by
`tan alpha=|(Im(z))/(Re(z))|=|(-1)/(-1)|=1 rArr alpha=(pi)/(4)`.
Clearly, the point representing the complex number z = -1 -i is P(-1, -1), which lies in the third quadrant.
`therefore" "arg(z) = theta = -(pi-alpha) = -(pi-(pi)/(4))=(-3pi)/(4)`.
Thus, `r = |z| = sqrt(2) and theta = (-3pi)/(4)`.
Hence, the required polar form of z = (-1 -i) is given by
`z=sqrt)2){cos((-3pi)/(4))+i sin((-3pi)/(4))}, i.e., sqrt(2)("cos"(3pi)/(4)+"i sin"(3pi)/(4))`


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