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-1628, convertinto\frac{-16}{1+i \sqrt{3}} polar form. |
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Answer» rationalising the denominator-16(1-i√3)/(1+i√3)(1-i√3)hence-16+16√3i/1+3hence-16/4+i4√3-4+4√3ihence in polar formfind thetahence theta is tan^-1(b/a)=tan^-1(-√3)hence theta is (-π/3)and r=√a^2+b^2r=√-4^2+(4√3)^2r=8hence in.polar form8(cos{-π/3)+isin(-π/3) |
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