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A complex equation

July 28, 2015 – 2:10 am

Find the number of solutions of the equation $z^{3}+\overline{z}=0$ .

Solution.

Given that $z^{3}+\overline{z}=0$ . Hence, $z^{3}=-z$ .

$|z|^{3} =|-\overline{z}| \Longrightarrow |z|^{3}=|z|$ .Hence, we get

$|z|(|z|-1)(|z|+1)=0 \Longrightarrow |z|=0, |z|=1$ (since $|z|+1>0$ )

If $|z|=1$ , we get $|z|^{2}=1 \Longrightarrow z.\overline{z}=1$ .

Thus, $z^{3}+\overline{z}=0 \Longrightarrow z^{3}+1/z=0$

Thus, $z^{4}+1=0 \Longrightarrow z^{4}=\cos{\pi}+i\sin{\pi}$ , that is,

$z=\cos{\frac{2k+1}{4}}\pi+i\sin{\frac{2k+1}{4}}\pi$ for $k=0,1,2,3$ . Therefore, the given equation has five solutions.

By Nalin Pithwa | Posted in Cnennai Math Institute Entrance Exam, Complex Numbers, IITJEE Mains, ISI Kolkatta Entrance Exam | Also tagged algebra for IITJEE | Comments (0)

Complex numbers

February 18, 2015 – 7:07 pm

Question: Prove that for any complex number z:

$|z+1| \geq \frac {1}{\sqrt {2}}$ or $|z^{2}+1| \geq 1$

Solution: Suppose by way of contradiction that

$|1+z| < \frac {1}{\sqrt {2}}$ and $|1+z^{2}| <1$

Setting $z=a+ib$ with $a,b \in R$ yields $z^{2}=a^{2}-b^{2}+i2ab$ .

We obtain

$(1+a^{2}-b^{2})^{2}+4a^{2}b^{2}<1$ and $(1+a)^{2}+b^{2}<1/2$

and consequently,

$(a^{2}+b^{2})^{2}+2(a^{2}-b^{2})<0$ and $2(a^{2}+b^{2})+4a+1<0$ .

Summing these inequalities implies $(a^{2}+b^{2})^{2}+(2a+1)^{2}>0$ which is a contradiction.

More later,

Nalin Pithwa

By Nalin Pithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Comments (0)