If P is the point in the Argand diagram corre | vedclass

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Question

(A) Let $z = \sqrt{3}+i$. The modulus is $|z| = \sqrt{(\sqrt{3})^2 + 1^2} = 2$ and the argument is $\arg(z) = 	an^{-1}(\frac{1}{\sqrt{3}}) = 30^{\cir

Vedclass · Accepted Answer

$-1+i\sqrt{3}$ or $1-i\sqrt{3}$

Vedclass · Answer

(A) Let $z = \sqrt{3}+i$. The modulus is $|z| = \sqrt{(\sqrt{3})^2 + 1^2} = 2$ and the argument is $\arg(z) = 	an^{-1}(\frac{1}{\sqrt{3}}) = 30^{\circ}$.Since $OPQ$ is an isosceles right-angled triangle with $\angle POQ = 90^{\circ}$ and $OP = OQ$,the point $Q$ can be obtained by rotating $P$ by $90^{\circ}$ in either the clockwise or counter-clockwise direction.If we rotate by $+90^{\circ}$,the argument of $Q$ is $30^{\circ} + 90^{\circ} = 120^{\circ}$. The complex number is $2(\cos 120^{\circ} + i\sin 120^{\circ}) = 2(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = -1 + i\sqrt{3}$.If we rotate by $-90^{\circ}$,the argument of $Q$ is $30^{\circ} - 90^{\circ} = -60^{\circ}$. The complex number is $2(\cos(-60^{\circ}) + i\sin(-60^{\circ})) = 2(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 1 - i\sqrt{3}$.Thus,$Q$ represents $-1+i\sqrt{3}$ or $1-i\sqrt{3}$.

If $P$ is the point in the Argand diagram corresponding to the complex number $\sqrt{3}+i$ and if $OPQ$ is an isosceles right-angled triangle,right-angled at $O$,then $Q$ represents the complex number

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