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Exercise 1.3 · Q106

Q.Find the modulus and argument of the complex number 1+2i1−3i\dfrac{1+2i}{1-3i}.

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dfrac1+2i1−3i=dfrac(1+2i)(1+3i)(1−3i)(1+3i)=dfrac1+3i+2i−61+9=dfrac−5+5i10=−dfrac12+dfrac12i\\dfrac{1+2i}{1-3i}=\\dfrac{(1+2i)(1+3i)}{(1-3i)(1+3i)}=\\dfrac{1+3i+2i-6}{1+9}=\\dfrac{-5+5i}{10}=-\\dfrac12+\\dfrac12i (same simplification as item 4(v) above). So ∣z∣=sqrttfrac14+tfrac14=dfracsqrt22|z|=\\sqrt{\\tfrac14+\\tfrac14}=\\dfrac{\\sqrt2}{2}, and since a=−tfrac12<0,b=tfrac12>0a=-\\tfrac12<0,b=\\tfrac12>0 (Quadrant II) …

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