Let OO be the origin, the point AA be z1=3+22iz_1 = \sqrt{3} + 2 \sqrt{2} i, the point B(z2)B(z_2) be such that 3∣z2∣=∣z1∣\sqrt{3} |z_2| = |z_1| and arg⁡(z2)=arg⁡(z1)+π2\arg(z_2) = \arg(z_1) + \frac{\pi}{2}. Then

Answer: z2=−223+iz_2 = -\frac{2\sqrt{2}}{\sqrt{3}} + i

Step-by-step solution

Step 1: Calculate Modulus and Argument of z1z_1

First, we need to find the modulus and argument of the complex number z1z_1. The modulus of a complex number a+bia+bi is given by a2+b2\sqrt{a^2+b^2}, and its argument is tan⁡−1(b/a)\tan^{-1}(b/a). For z1=3+22iz_1 = \sqrt{3} + 2\sqrt{2}i, we calculate its modulus as (3)2+(22)2=3+8=11\sqrt{(\sqrt{3})^2 + (2\sqrt{2})^2} = \sqrt{3+8} = \sqrt{11}. The argument is tan⁡−1(223)\tan^{-1}\left(\frac{2\sqrt{2}}{\sqrt{3}}\right).

Step 2: Calculate Modulus of z2z_2

We are given the relation 3∣z2∣=∣z1∣\sqrt{3} |z_2| = |z_1|. We can use the modulus of z1z_1 we just calculated to find the modulus of z2z_2. Dividing both sides by 3\sqrt{3}, we get ∣z2∣=113|z_2| = \frac{\sqrt{11}}{\sqrt{3}}, which simplifies to 113\sqrt{\frac{11}{3}}.

Step 3: Calculate Argument of z2z_2

We are given that the argument of z2z_2 is π2\frac{\pi}{2} radians (or 90 degrees) greater than the argument of z1z_1. This means that the vector representing z2z_2 is obtained by rotating the vector representing z1z_1 counter-clockwise by 90 degrees.

Step 4: Express z2z_2 in terms of z1z_1

We can express z2z_2 in polar form using its modulus and argument. Since arg⁡(z2)=arg⁡(z1)+π2\arg(z_2) = \arg(z_1) + \frac{\pi}{2}, we can write eiarg⁡(z2)=eiarg⁡(z1)eiπ2e^{i \arg(z_2)} = e^{i \arg(z_1)} e^{i \frac{\pi}{2}}. Also, we know that eiπ2=cos⁡(π2)+isin⁡(π2)=0+i(1)=ie^{i \frac{\pi}{2}} = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = 0 + i(1) = i. Substituting these into the polar form for z2z_2, we get z2=∣z1∣3eiarg⁡(z1)iz_2 = \frac{|z_1|}{\sqrt{3}} e^{i \arg(z_1)} i. Since z1=∣z1∣eiarg⁡(z1)z_1 = |z_1| e^{i \arg(z_1)}, we can write z2=i3z1z_2 = \frac{i}{\sqrt{3}} z_1.

Step 5: Substitute z1z_1 to find z2z_2

Now we substitute the given value of z1=3+22iz_1 = \sqrt{3} + 2\sqrt{2}i into the expression for z2z_2. We multiply i3\frac{i}{\sqrt{3}} by z1z_1. Remember that i2=−1i^2 = -1. After distributing and simplifying, we get z2=−223+iz_2 = -\frac{2\sqrt{2}}{\sqrt{3}} + i.

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