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Question Number 215417 by MATHEMATICSAM last updated on 06/Jan/25

$$a\mathrm{and}b\mathrm{are}\mathrm{complex}\mathrm{numbers}\mathrm{such}\mathrm{that}$$$$\mid b\mid =1.\mathrm{Find}\mid \frac{b-a}{1-\overline{ab}}\mid$$

Answered by MrGaster last updated on 06/Jan/25

$$\mathrm{Let}a=x+yi,b=\cos \theta +i\sin \theta$$$$b-a=(\cos \theta -x)+i(\sin \theta -y)$$$$1=\overline{ab}=1-(x+iy)(\cos \theta -i\sin \theta )$$$$=1-(x\cos \theta +y\sin \theta )-i(x\sin \theta -y\cos \theta )$$$$\mid b-a\mid =\sqrt{{(\cos \theta -x)}^2+{(\sin \theta -y)}^2}$$$$\mid 1-\overline{ab}\mid =\sqrt{{1-x\cos \theta -y\sin \theta )}^2+{(x\sin \theta -y\cos \theta )}^2}$$$$\mid \frac{b-a}{1-\overline{ab}}\mid =\frac{\sqrt{{(\cos \theta -x)}^2+{(\sin \theta -y)}^2}}{\sqrt{{(1-x\cos \theta -y\sin \theta )}^2+{(x\sin \theta -y\cos \theta )}^2}}$$$$=1$$$$\mid \frac{b-a}{1-\overline{ab}}\mid =1✓$$

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