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JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 62

Let z1=1+2iz_1=1+2i, z2=2+iz_2=2+i, 1z1+1z2=6w\dfrac1{z_1}+\dfrac1{z_2}=\dfrac6w, and z=iw2−wˉz=\dfrac{iw}{2-\bar w}.

Consider the statements

(S1_1) ∣z∣=517|z|=\dfrac5{\sqrt{17}}

(S2_2) arg⁡(z)+arg⁡(w)=tan⁡−1(53)\arg(z)+\arg(w)=\tan^{-1}\left(\dfrac53\right).

Then

Choose one option
Answer and worked solution

Official answer: D

w. 1z1+1z2=3−3i5\frac1{z_1}+\frac1{z_2}=\frac{3-3i}5, so w=303−3i=5(1+i)w=\frac{30}{3-3i}=5(1+i).

z =5i(1+i)2−5+5i=5(−1+i)−3+5i=\dfrac{5i(1+i)}{2-5+5i}=\dfrac{5(-1+i)}{-3+5i}: ∣z∣=5234=517|z|=\dfrac{5\sqrt2}{\sqrt{34}}=\dfrac5{\sqrt{17}}.

Arguments. arg⁡z=3π4−(π−tan⁡−153)=tan⁡−153−π4\arg z=\frac{3\pi}4-\left(\pi-\tan^{-1}\frac53\right)=\tan^{-1}\frac53-\frac\pi4 and arg⁡w=π4\arg w=\frac\pi4, so the sum is tan⁡−153\tan^{-1}\frac53. Both correct.

(D)

Topics:Modulus and argument
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