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Exercise 2.6 · Q3

Q.Obtain the Cartesian form of the locus of z=x+iyz=x+iy in each of the following cases:

(i) [Re⁡(iz)]2=3[\operatorname{Re}(iz)]^2=3
(ii) Im⁡[(1−i)z+1]=0\operatorname{Im}[(1-i)z+1]=0
(iii) ∣z+i∣=∣z−1∣|z+i|=|z-1|
(iv) z‾=z−1\overline z=z^{-1}.
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Each part substitutes z=x+iyz=x+iy into the given condition and reduces it algebraically to a real Cartesian equation in x,yx,y; we take them one at a time.

Step 1. (i) [Re⁡(iz)]2=3[\operatorname{Re}(iz)]^2=3. With z=x+iyz=x+iy, iz=i(x+iy)=ix−y=−y+ixiz=i(x+iy)=ix-y=-y+ix, so Re⁡(iz)=−y\operatorname{Re}(iz)=-y. The condition becomes (−y)2=3⇒y2=3(-y)^2=3\Rightarrow y^2=3 (a pair of horizontal lines y=±3y=\pm\sqrt3).

Step 2. (ii) Im⁡[(1−i)z+1]=0\operatorname{Im}[(1-i)z+1]=0. Compute (1−i)z=(1−i)(x+iy)=x+iy−ix−i2y=(x+y)+i(y−x)(1-i)z=(1-i)(x+iy)=x+iy-ix-i^2y=(x+y)+i(y-x), so (1−i)z+1=(x+y+1)+i(y−x)(1-i)z+1=(x+y+1)+i(y-x). Its imaginary part is y−xy-x; setting this to 00 gives y−x=0y-x=0, i.e. x−y=0x-y=0.

Step 3. (iii) ∣z+i∣=∣z−1∣|z+i|=|z-1|. With z=x+iyz=x+iy: z+i=x+i(y+1)z+i=x+i(y+1) and z−1=(x−1)+iyz-1=(x-1)+iy. Equating moduli and squaring:

x2+(y+1)2=(x−1)2+y2 ⇒ x2+y2+2y+1=x2−2x+1+y2 ⇒ 2y=−2x ⇒ x+y=0.x^2+(y+1)^2=(x-1)^2+y^2\ \Rightarrow\ x^2+y^2+2y+1=x^2-2x+1+y^2\ \Rightarrow\ 2y=-2x\ \Rightarrow\ x+y=0. …

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