Tag Archives: complex numbers

A complex equation

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


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.

Complex numbers

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